{"id":543391,"date":"2021-09-30T04:24:24","date_gmt":"2021-09-30T04:24:24","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543391"},"modified":"2021-10-01T06:50:40","modified_gmt":"2021-10-01T06:50:40","slug":"rd-sharma-solutions-for-class-11-maths-chapter-12-mathematical-induction","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-12-mathematical-induction\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 12\u2013Mathematical Induction"},"content":{"rendered":"\n

Class 11: Maths Chapter 12 solutions. Complete Class 11 Maths Chapter 12 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 12\u2013Mathematical Induction<\/h2>\n\n\n\n

RD Sharma 11th Maths Chapter 12, Class 11 Maths Chapter 12 solutions<\/p>\n\n\n\n

EXERCISE 12.1 PAGE NO: 12.3<\/p>\n\n\n\n

1. If P (n) is the statement \u201cn (n + 1) is even\u201d, then what is P (3)?
Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

P (n) = n (n + 1) is even.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

P (3) = 3 (3 + 1)<\/p>\n\n\n\n

= 3 (4)<\/p>\n\n\n\n

= 12<\/p>\n\n\n\n

Hence, P (3) = 12, P (3) is also even.<\/p>\n\n\n\n

2. If P (n) is the statement \u201cn3<\/sup> + n is divisible by 3\u201d, prove that P (3) is true but P (4) is not true.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

P (n) = n3<\/sup> + n is divisible by 3<\/p>\n\n\n\n

We have P (n) = n3<\/sup> + n<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

P (3) = 33<\/sup> + 3<\/p>\n\n\n\n

= 27 + 3<\/p>\n\n\n\n

= 30<\/p>\n\n\n\n

P (3) = 30, So it is divisible by 3<\/p>\n\n\n\n

Now, let\u2019s check with P (4)<\/p>\n\n\n\n

P (4) = 43<\/sup> + 4<\/p>\n\n\n\n

= 64 + 4<\/p>\n\n\n\n

= 68<\/p>\n\n\n\n

P (4) = 68, so it is not divisible by 3<\/p>\n\n\n\n

Hence, P (3) is true and P (4) is not true.<\/p>\n\n\n\n

3. If P (n) is the statement \u201c2n<\/sup> \u2265 3n\u201d, and if P (r) is true, prove that P (r + 1) is true.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

P (n) = \u201c2n<\/sup> \u2265 3n\u201d and p(r) is true.<\/p>\n\n\n\n

We have, P (n) = 2n<\/sup> \u2265 3n<\/p>\n\n\n\n

Since, P (r) is true<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2r<\/sup>\u2265 3r<\/p>\n\n\n\n

Now, let\u2019s multiply both sides by 2<\/p>\n\n\n\n

2\u00d72r<\/sup>\u2265 3r\u00d72<\/p>\n\n\n\n

2r + 1<\/sup>\u2265 6r<\/p>\n\n\n\n

2r + 1<\/sup>\u2265 3r + 3r [since 3r>3 = 3r + 3r\u22653 + 3r]<\/p>\n\n\n\n

\u2234 2r + 1<\/sup>\u2265 3(r + 1)<\/p>\n\n\n\n

Hence, P (r + 1) is true.<\/p>\n\n\n\n

4. If P (n) is the statement \u201cn2<\/sup> + n\u201d is even\u201d, and if P (r) is true, then P (r + 1) is true
Solution:
<\/strong>
Given:<\/p>\n\n\n\n

P (n) = n2<\/sup> + n is even and P (r) is true, then r2<\/sup> + r is even<\/p>\n\n\n\n

Let us consider r2<\/sup> + r = 2k \u2026 (i)<\/p>\n\n\n\n

Now, (r + 1)2<\/sup> + (r + 1)<\/p>\n\n\n\n

r2<\/sup> + 1 + 2r + r + 1<\/p>\n\n\n\n

(r2<\/sup> + r) + 2r + 2<\/p>\n\n\n\n

2k + 2r + 2 [from equation (i)]<\/p>\n\n\n\n

2(k + r + 1)<\/p>\n\n\n\n

2\u03bc<\/p>\n\n\n\n

\u2234 (r + 1)2<\/sup> + (r + 1) is Even.<\/p>\n\n\n\n

Hence, P (r + 1) is true.<\/p>\n\n\n\n

5. Given an example of a statement P (n) such that it is true for all n \u03f5 N.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider<\/p>\n\n\n\n

P (n) = 1 + 2 + 3 + \u2013 \u2013 \u2013 \u2013 \u2013 + n = n(n+1)\/2<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

P (n) is true for all natural numbers.<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

6. If P (n) is the statement \u201cn2<\/sup> \u2013 n + 41 is prime\u201d, prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
Solution:
<\/strong>
Given:<\/p>\n\n\n\n

P(n) = n2<\/sup> \u2013 n + 41 is prime.<\/p>\n\n\n\n

P(n) = n2<\/sup> \u2013 n + 41<\/p>\n\n\n\n

P (1) = 1 \u2013 1 + 41<\/p>\n\n\n\n

= 41<\/p>\n\n\n\n

P (1) is Prime.<\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

P(2) = 22<\/sup> \u2013 2 + 41<\/p>\n\n\n\n

= 4 \u2013 2 + 41<\/p>\n\n\n\n

= 43<\/p>\n\n\n\n

P (2) is prime.<\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

P (3) = 32<\/sup> \u2013 3 + 41<\/p>\n\n\n\n

= 9 \u2013 3 + 41<\/p>\n\n\n\n

= 47<\/p>\n\n\n\n

P (3) is prime<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

P (41) = (41)2<\/sup> \u2013 41 + 41<\/p>\n\n\n\n

= 1681<\/p>\n\n\n\n

P (41) is not prime<\/p>\n\n\n\n

Hence, P (1), P(2), P (3) are true but P (41) is not true.<\/p>\n\n\n\n

EXERCISE 12.2 PAGE NO: 12.27<\/p>\n\n\n\n

Prove the following by the principle of mathematical induction:<\/strong><\/p>\n\n\n\n

1. 1 + 2 + 3 + \u2026 + n = n (n +1)\/2 i.e., the sum of the first n natural numbers is n (n + 1)\/2.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider P (n) = 1 + 2 + 3 + \u2026.. + n = n (n +1)\/2<\/p>\n\n\n\n

For, n = 1<\/p>\n\n\n\n

LHS of P (n) = 1<\/p>\n\n\n\n

RHS of P (n) = 1 (1+1)\/2 = 1<\/p>\n\n\n\n

So, LHS = RHS<\/p>\n\n\n\n

Since, P (n) is true for n = 1<\/p>\n\n\n\n

Let us consider P (n) be the true for n = k, so<\/p>\n\n\n\n

1 + 2 + 3 + \u2026. + k = k (k+1)\/2 \u2026 (i)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

(1 + 2 + 3 + \u2026 + k) + (k + 1) = k (k+1)\/2 + (k+1)<\/p>\n\n\n\n

= (k + 1) (k\/2 + 1)<\/p>\n\n\n\n

= [(k + 1) (k + 2)] \/ 2<\/p>\n\n\n\n

= [(k+1) [(k+1) + 1]] \/ 2<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

P (n) is true for all n \u2208 N<\/p>\n\n\n\n

So, by the principle of Mathematical Induction<\/p>\n\n\n\n

Hence, P (n) = 1 + 2 + 3 + \u2026.. + n = n (n +1)\/2 is true for all n \u2208 N.<\/p>\n\n\n\n

2. 12<\/sup> + 22<\/sup> + 32<\/sup> + \u2026 + n2<\/sup> = [n (n+1) (2n+1)]\/6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider P (n) = 12<\/sup> + 22<\/sup> + 32<\/sup> + \u2026 + n2<\/sup> = [n (n+1) (2n+1)]\/6<\/p>\n\n\n\n

For, n = 1<\/p>\n\n\n\n

P (1) = [1 (1+1) (2+1)]\/6<\/p>\n\n\n\n

1 = 1<\/p>\n\n\n\n

P (n) is true for n = 1<\/p>\n\n\n\n

Let P (n) is true for n = k, so<\/p>\n\n\n\n

P (k): 12<\/sup> + 22<\/sup> + 32<\/sup> + \u2026 + k2<\/sup> = [k (k+1) (2k+1)]\/6<\/p>\n\n\n\n

Let\u2019s check for P (n) = k + 1, so<\/p>\n\n\n\n

P (k) = 12<\/sup> + 22<\/sup> + 32<\/sup> + \u2013 \u2013 \u2013 \u2013 \u2013 + k2<\/sup> + (k + 1)2<\/sup> = [k + 1 (k+2) (2k+3)] \/6<\/p>\n\n\n\n

= 12<\/sup> + 22<\/sup> + 32<\/sup> + \u2013 \u2013 \u2013 \u2013 \u2013 + k2<\/sup> + (k + 1)2<\/sup><\/p>\n\n\n\n

= [k + 1 (k+2) (2k+3)] \/6 + (k + 1)2<\/sup><\/p>\n\n\n\n

= (k +1) [(2k2<\/sup> + k)\/6 + (k + 1)\/1]<\/p>\n\n\n\n

= (k +1) [2k2<\/sup> + k + 6k + 6]\/6<\/p>\n\n\n\n

= (k +1) [2k2<\/sup> + 7k + 6]\/6<\/p>\n\n\n\n

= (k +1) [2k2<\/sup> + 4k + 3k + 6]\/6<\/p>\n\n\n\n

= (k +1) [2k(k + 2) + 3(k + 2)]\/6<\/p>\n\n\n\n

= [(k +1) (2k + 3) (k + 2)] \/ 6<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

3. 1 + 3 + 32<\/sup> + \u2026 + 3n-1<\/sup> = (3n<\/sup> \u2013 1)\/2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1 + 3 + 32<\/sup> + \u2013 \u2013 \u2013 \u2013 + 3n \u2013 1<\/sup> = (3n<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

Now, For n = 1<\/p>\n\n\n\n

P (1) = 1 = (31<\/sup> \u2013 1)\/2 = 2\/2 =1<\/p>\n\n\n\n

P (n) is true for n = 1<\/p>\n\n\n\n

Now, let\u2019s check for P (n) is true for n = k<\/p>\n\n\n\n

P (k) = 1 + 3 + 32<\/sup> + \u2013 \u2013 \u2013 \u2013 + 3k \u2013 1<\/sup> = (3k<\/sup> \u2013 1)\/2 \u2026 (i)<\/p>\n\n\n\n

Now, we have to show P (n) is true for n = k + 1<\/p>\n\n\n\n

P (k + 1) = 1 + 3 + 32<\/sup> + \u2013 \u2013 \u2013 \u2013 + 3k<\/sup> = (3k+1<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

Then, {1 + 3 + 32<\/sup> + \u2013 \u2013 \u2013 \u2013 + 3k \u2013 1<\/sup>} + 3k + 1 \u2013 1<\/sup><\/p>\n\n\n\n

= (3k \u2013 1)\/2 + 3k<\/sup> using equation (i)<\/p>\n\n\n\n

= (3k \u2013 1 + 2\u00d73k<\/sup>)\/2<\/p>\n\n\n\n

= (3\u00d73 k \u2013 1)\/2<\/p>\n\n\n\n

= (3k+1<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

4. 1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/n(n+1) = n\/(n+1)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/n(n+1) = n\/(n+1)<\/p>\n\n\n\n

For, n = 1<\/p>\n\n\n\n

P (n) = 1\/1.2 = 1\/1+1<\/p>\n\n\n\n

1\/2 = 1\/2<\/p>\n\n\n\n

P (n) is true for n = 1<\/p>\n\n\n\n

Let\u2019s check for P (n) is true for n = k,<\/p>\n\n\n\n

1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/k(k+1) + k\/(k+1) (k+2) = (k+1)\/(k+2)<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/k(k+1) + k\/(k+1) (k+2)<\/p>\n\n\n\n

= 1\/(k+1)\/(k+2) + k\/(k+1)<\/p>\n\n\n\n

= 1\/(k+1) [k(k+2)+1]\/(k+2)<\/p>\n\n\n\n

= 1\/(k+1) [k2<\/sup> + 2k + 1]\/(k+2)<\/p>\n\n\n\n

=1\/(k+1) [(k+1) (k+1)]\/(k+2)<\/p>\n\n\n\n

= (k+1) \/ (k+2)<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

5. 1 + 3 + 5 + \u2026 + (2n \u2013 1) = n2<\/sup> i.e., the sum of first n odd natural numbers is n2<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 1 + 3 + 5 + \u2026 + (2n \u2013 1) = n2<\/sup><\/p>\n\n\n\n

Let us check P (n) is true for n = 1<\/p>\n\n\n\n

P (1) = 1 =12<\/sup><\/p>\n\n\n\n

1 = 1<\/p>\n\n\n\n

P (n) is true for n = 1<\/p>\n\n\n\n

Now, Let\u2019s check P (n) is true for n = k<\/p>\n\n\n\n

P (k) = 1 + 3 + 5 + \u2026 + (2k \u2013 1) = k2<\/sup> \u2026 (i)<\/p>\n\n\n\n

We have to show that<\/p>\n\n\n\n

1 + 3 + 5 + \u2026 + (2k \u2013 1) + 2(k + 1) \u2013 1 = (k + 1)2<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

1 + 3 + 5 + \u2026 + (2k \u2013 1) + 2(k + 1) \u2013 1<\/p>\n\n\n\n

= k2<\/sup> + (2k + 1)<\/p>\n\n\n\n

= k2<\/sup> + 2k + 1<\/p>\n\n\n\n

= (k + 1)2<\/sup><\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

6. 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3n-1) (3n+2) = n\/(6n+4)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3n-1) (3n+2) = n\/(6n+4)<\/p>\n\n\n\n

Let us check P (n) is true for n = 1<\/p>\n\n\n\n

P (1): 1\/2.5 = 1\/6.1+4 => 1\/10 = 1\/10<\/p>\n\n\n\n

P (1) is true.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Let us check for P (k) is true, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3k-1) (3k+2) = k\/(6k+4)<\/p>\n\n\n\n

P (k +1): 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3k-1)(3k+2) + 1\/(3k+3-1)(3k+3+2)<\/p>\n\n\n\n

: k\/(6k+4) + 1\/(3k+2)(3k+5)<\/p>\n\n\n\n

: [k(3k+5)+2] \/ [2(3k+2)(3k+5)]<\/p>\n\n\n\n

: (k+1) \/ (6(k+1)+4)<\/p>\n\n\n\n

P (k + 1) is true.<\/p>\n\n\n\n

Hence proved by mathematical induction.<\/p>\n\n\n\n

7. 1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3n-2)(3n+1) = n\/3n+1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3n-2)(3n+1) = n\/3n+1<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1\/1.4 = 1\/4<\/p>\n\n\n\n

1\/4 = 1\/4<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k) = 1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3k-2)(3k+1) = k\/3k+1 \u2026 (i)<\/p>\n\n\n\n

So,[1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3k-2)(3k+1)]+ 1\/(3k+1)(3k+4)<\/p>\n\n\n\n

= k\/(3k+1) + 1\/(3k+1)(3k+4)<\/p>\n\n\n\n

= 1\/(3k+1) [k\/1 + 1\/(3k+4)]<\/p>\n\n\n\n

= 1\/(3k+1) [k(3k+4)+1]\/(3k+4)<\/p>\n\n\n\n

= 1\/(3k+1) [3k2<\/sup> + 4k + 1]\/ (3k+4)<\/p>\n\n\n\n

= 1\/(3k+1) [3k2<\/sup> + 3k+k+1]\/(3k+4)<\/p>\n\n\n\n

= [3k(k+1) + (k+1)] \/ [(3k+4) (3k+1)]<\/p>\n\n\n\n

= [(3k+1)(k+1)] \/ [(3k+4) (3k+1)]<\/p>\n\n\n\n

= (k+1) \/ (3k+4)<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

8. 1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2n+1)(2n+3) = n\/3(2n+3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2n+1)(2n+3) = n\/3(2n+3)<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1\/3.5 = 1\/3(2.1+3)<\/p>\n\n\n\n

: 1\/15 = 1\/15<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k) = 1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2k+1)(2k+3) = k\/3(2k+3) \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2k+1)(2k+3) + 1\/[2(k+1)+1][2(k+1)+3]<\/p>\n\n\n\n

1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2k+1)(2k+3) + 1\/(2k+3)(2k+5)<\/p>\n\n\n\n

Now substituting the value of P (k) we get,<\/p>\n\n\n\n

= k\/3(2k+3) + 1\/(2k+3)(2k+5)<\/p>\n\n\n\n

= [k(2k+5)+3] \/ [3(2k+3)(2k+5)]<\/p>\n\n\n\n

= (k+1) \/ [3(2(k+1)+3)]<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

9. 1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4n-1)(4n+3) = n\/3(4n+3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4n-1)(4n+3) = n\/3(4n+3)<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1\/3.7 = 1\/(4.1-1)(4+3)<\/p>\n\n\n\n

: 1\/21 = 1\/21<\/p>\n\n\n\n

P (n) is true for n =1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4k-1)(4k+3) = k\/3(4k+3) \u2026. (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4k-1)(4k+3) + 1\/(4k+3)(4k+7)<\/p>\n\n\n\n

Substituting the value of P (k) we get,<\/p>\n\n\n\n

= k\/(4k+3) + 1\/(4k+3)(4k+7)<\/p>\n\n\n\n

= 1\/(4k+3) [k(4k+7)+3] \/ [3(4k+7)]<\/p>\n\n\n\n

= 1\/(4k+3) [4k2<\/sup> + 7k +3]\/ [3(4k+7)]<\/p>\n\n\n\n

= 1\/(4k+3) [4k2<\/sup> + 3k+4k+3] \/ [3(4k+7)]<\/p>\n\n\n\n

= 1\/(4k+3) [4k(k+1)+3(k+1)]\/ [3(4k+7)]<\/p>\n\n\n\n

= 1\/(4k+3) [(4k+3)(k+1)] \/ [3(4k+7)]<\/p>\n\n\n\n

= (k+1) \/ [3(4k+7)]<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

10. 1.2 + 2.22<\/sup> + 3.23<\/sup> + \u2026 + n.2n <\/sup>= (n\u20131) 2n + 1<\/sup> + 2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 1.2 + 2.22<\/sup> + 3.23<\/sup> + \u2026 + n.2n <\/sup>= (n\u20131) 2n + 1<\/sup> + 2<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1):1.2 = 0.20<\/sup> + 2<\/p>\n\n\n\n

: 2 = 2<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 1.2 + 2.22<\/sup> + 3.23<\/sup> + \u2026 + k.2k <\/sup>= (k\u20131) 2k + 1<\/sup> + 2 \u2026. (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

{1.2 + 2.22<\/sup> + 3.23<\/sup> + \u2026 + k.2k<\/sup>} + (k + 1)2k + 1<\/sup><\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= [(k \u2013 1)2k + 1<\/sup> + 2] + (k + 1)2k + 1<\/sup> using equation (i)<\/p>\n\n\n\n

= (k \u2013 1)2k + 1<\/sup> + 2 + (k + 1)2k + 1<\/sup><\/p>\n\n\n\n

= 2k + 1<\/sup>(k \u2013 1 + k + 1) + 2<\/p>\n\n\n\n

= 2k + 1<\/sup> \u00d7 2k + 2<\/p>\n\n\n\n

= k \u00d7 2k + 2<\/sup> + 2<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

11. 2 + 5 + 8 + 11 + \u2026 + (3n \u2013 1) = 1\/2 n (3n + 1)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n) = 2 + 5 + 8 + 11 + \u2026 + (3n \u2013 1) = 1\/2 n (3n + 1)<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 2 = 1\/2 \u00d7 1 \u00d7 4<\/p>\n\n\n\n

: 2 = 2<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k) = 2 + 5 + 8 + 11 + \u2026 + (3k \u2013 1) = 1\/2 k (3k + 1) \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2 + 5 + 8 + 11 + \u2026 + (3k \u2013 1) + (3k + 2)<\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= 1\/2 \u00d7 k (3k + 1) + (3k + 2) by using equation (i)<\/p>\n\n\n\n

= [3k2<\/sup> + k + 2 (3k + 2)] \/ 2<\/p>\n\n\n\n

= [3k2<\/sup> + k + 6k + 2] \/ 2<\/p>\n\n\n\n

= [3k2<\/sup> + 7k + 2] \/ 2<\/p>\n\n\n\n

= [3k2<\/sup> + 4k + 3k + 2] \/ 2<\/p>\n\n\n\n

= [3k (k+1) + 4(k+1)] \/ 2<\/p>\n\n\n\n

= [(k+1) (3k+4)] \/2<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

12. 1.3 + 2.4 + 3.5 + \u2026 + n. (n+2) = 1\/6 n (n+1) (2n+7)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 1.3 + 2.4 + 3.5 + \u2026 + n. (n+2) = 1\/6 n (n+1) (2n+7)<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1.3 = 1\/6 \u00d7 1 \u00d7 2 \u00d7 9<\/p>\n\n\n\n

: 3 = 3<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 1.3 + 2.4 + 3.5 + \u2026 + k. (k+2) = 1\/6 k (k+1) (2k+7) \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1.3 + 2.4 + 3.5 + \u2026 + k. (k+2) + (k+1) (k+3)<\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= 1\/6 k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)<\/p>\n\n\n\n

= (k+1) [{k(2k+7)\/6} + {(k+3)\/1}]<\/p>\n\n\n\n

= (k+1) [(2k2<\/sup> + 7k + 6k + 18)] \/ 6<\/p>\n\n\n\n

= (k+1) [2k2<\/sup> + 13k + 18] \/ 6<\/p>\n\n\n\n

= (k+1) [2k2<\/sup> + 9k + 4k + 18] \/ 6<\/p>\n\n\n\n

= (k+1) [2k(k+2) + 9(k+2)] \/ 6<\/p>\n\n\n\n

= (k+1) [(2k+9) (k+2)] \/ 6<\/p>\n\n\n\n

= 1\/6 (k+1) (k+2) (2k+9)<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

13. 1.3 + 3.5 + 5.7 + \u2026 + (2n \u2013 1) (2n + 1) = n(4n2<\/sup> + 6n \u2013 1)\/3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 1.3 + 3.5 + 5.7 + \u2026 + (2n \u2013 1) (2n + 1) = n(4n2<\/sup> + 6n \u2013 1)\/3<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): (2.1 \u2013 1) (2.1 + 1) = 1(4.12<\/sup> + 6.1 -1)\/3<\/p>\n\n\n\n

: 1\u00d73 = 1(4+6-1)\/3<\/p>\n\n\n\n

: 3 = 3<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 1.3 + 3.5 + 5.7 + \u2026 + (2k \u2013 1) (2k + 1) = k(4k2<\/sup> + 6k \u2013 1)\/3 \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1.3 + 3.5 + 5.7 + \u2026 + (2k \u2013 1) (2k + 1) + (2k + 1) (2k + 3)<\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= k(4k2<\/sup> + 6k \u2013 1)\/3 + (2k + 1) (2k + 3) by using equation (i)<\/p>\n\n\n\n

= [k(4k2<\/sup> + 6k-1) + 3 (4k2<\/sup> + 6k + 2k + 3)] \/ 3<\/p>\n\n\n\n

= [4k3<\/sup> + 6k2<\/sup> \u2013 k + 12k2<\/sup> + 18k + 6k + 9] \/3<\/p>\n\n\n\n

= [4k3<\/sup> + 18k2<\/sup> + 23k + 9] \/3<\/p>\n\n\n\n

= [4k3<\/sup> + 4k2<\/sup> + 14k2<\/sup> + 14k +9k + 9] \/3<\/p>\n\n\n\n

= [(k+1) (4k2<\/sup> + 8k +4 + 6k + 6 \u2013 1)] \/ 3<\/p>\n\n\n\n

= [(k+1) 4[(k+1)2<\/sup> + 6(k+1) -1]] \/3<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

14. 1.2 + 2.3 + 3.4 + \u2026 + n(n+1) = [n (n+1) (n+2)] \/ 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 1.2 + 2.3 + 3.4 + \u2026 + n(n+1) = [n (n+1) (n+2)] \/ 3<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1(1+1) = [1(1+1) (1+2)] \/3<\/p>\n\n\n\n

: 2 = 2<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 1.2 + 2.3 + 3.4 + \u2026 + k(k+1) = [k (k+1) (k+2)] \/ 3 \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1.2 + 2.3 + 3.4 + \u2026 + k(k+1) + (k+1) (k+2)<\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= [k (k+1) (k+2)] \/ 3 + (k+1) (k+2) by using equation (i)<\/p>\n\n\n\n

= (k+2) (k+1) [k\/2 + 1]<\/p>\n\n\n\n

= [(k+1) (k+2) (k+3)] \/3<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

15. 1\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2n<\/sup> = 1 \u2013 1\/2n<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 1\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2n<\/sup> = 1 \u2013 1\/2n<\/sup><\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1\/21<\/sup> = 1 \u2013 1\/21<\/sup><\/p>\n\n\n\n

: 1\/2 = 1\/2<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

Let P (k): 1\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2k<\/sup> = 1 \u2013 1\/2k<\/sup> \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2k<\/sup> + 1\/2k+1<\/sup><\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= 1 \u2013 1\/2k <\/sup>+ 1\/2k+1<\/sup> by using equation (i)<\/p>\n\n\n\n

= 1 \u2013 ((2-1)\/2k+1<\/sup>)<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

16. 12<\/sup> + 32<\/sup> + 52<\/sup> + \u2026 + (2n \u2013 1)2<\/sup> = 1\/3 n (4n2<\/sup> \u2013 1)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 12<\/sup> + 32<\/sup> + 52<\/sup> + \u2026 + (2n \u2013 1)2<\/sup> = 1\/3 n (4n2<\/sup> \u2013 1)<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): (2.1 \u2013 1)2<\/sup> = 1\/3 \u00d7 1 \u00d7 (4 \u2013 1)<\/p>\n\n\n\n

: 1 = 1<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 12<\/sup> + 32<\/sup> + 52<\/sup> + \u2026 + (2k \u2013 1)2<\/sup> = 1\/3 k (4k2<\/sup> \u2013 1) \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

12<\/sup> + 32<\/sup> + 52<\/sup> + \u2026 + (2k \u2013 1)2<\/sup> + (2k + 1)2<\/sup><\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= 1\/3 k (4k2<\/sup> \u2013 1) + (2k + 1)2<\/sup> by using equation (i)<\/p>\n\n\n\n

= 1\/3 k (2k + 1) (2k \u2013 1) + (2k + 1)2<\/sup><\/p>\n\n\n\n

= (2k + 1) [{k(2k-1)\/3} + (2k+1)]<\/p>\n\n\n\n

= (2k + 1) [2k2<\/sup> \u2013 k + 3(2k+1)] \/ 3<\/p>\n\n\n\n

= (2k + 1) [2k2<\/sup> \u2013 k + 6k + 3] \/ 3<\/p>\n\n\n\n

= [(2k+1) 2k2<\/sup> + 5k + 3] \/3<\/p>\n\n\n\n

= [(2k+1) (2k(k+1)) + 3 (k+1)] \/3<\/p>\n\n\n\n

= [(2k+1) (2k+3) (k+1)] \/3<\/p>\n\n\n\n

= (k+1)\/3 [4k2<\/sup> + 6k + 2k + 3]<\/p>\n\n\n\n

= (k+1)\/3 [4k2<\/sup> + 8k \u2013 1]<\/p>\n\n\n\n

= (k+1)\/3 [4(k+1)2<\/sup> \u2013 1]<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

17. a + ar + ar2<\/sup> + \u2026 + arn \u2013 1<\/sup> = a [(rn<\/sup> \u2013 1)\/(r \u2013 1)], r \u2260 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): a + ar + ar2<\/sup> + \u2026 + arn \u2013 1<\/sup> = a [(rn<\/sup> \u2013 1)\/(r \u2013 1)]<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): a = a (r1<\/sup> \u2013 1)\/(r-1)<\/p>\n\n\n\n

: a = a<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): a + ar + ar2<\/sup> + \u2026 + ark \u2013 1<\/sup> = a [(rk<\/sup> \u2013 1)\/(r \u2013 1)] \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a + ar + ar2<\/sup> + \u2026 + ark \u2013 1<\/sup> + ark<\/sup><\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= a [(rk<\/sup> \u2013 1)\/(r \u2013 1)] + ark<\/sup> by using equation (i)<\/p>\n\n\n\n

= a[rk<\/sup> \u2013 1 + rk<\/sup>(r-1)] \/ (r-1)<\/p>\n\n\n\n

= a[rk<\/sup> \u2013 1 + rk+1<\/sup> \u2013 r\u2011k<\/sup>] \/ (r-1)<\/p>\n\n\n\n

= a[rk+1<\/sup> \u2013 1] \/ (r-1)<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

18. a + (a + d) + (a + 2d) + \u2026 + (a + (n-1)d) = n\/2 [2a + (n-1)d]<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): a + (a + d) + (a + 2d) + \u2026 + (a + (n-1)d) = n\/2 [2a + (n-1)d]<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): a = \u00bd [2a + (1-1)d]<\/p>\n\n\n\n

: a = a<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): a + (a + d) + (a + 2d) + \u2026 + (a + (k-1)d) = k\/2 [2a + (k-1)d] \u2026 (i)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a + (a + d) + (a + 2d) + \u2026 + (a + (k-1)d) + (a + (k)d)<\/p>\n\n\n\n

Now, substituting the value of P (k) we get,<\/p>\n\n\n\n

= k\/2 [2a + (k-1)d] + (a + kd) by using equation (i)<\/p>\n\n\n\n

= [2ka + k(k-1)d + 2(a+kd)] \/ 2<\/p>\n\n\n\n

= [2ka + k2<\/sup>d \u2013 kd + 2a + 2kd] \/ 2<\/p>\n\n\n\n

= [2ka + 2a + k2<\/sup>d + kd] \/ 2<\/p>\n\n\n\n

= [2a(k+1) + d(k2<\/sup> + k)] \/ 2<\/p>\n\n\n\n

= (k+1)\/2 [2a + kd]<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

19. 52n<\/sup> \u2013 1 is divisible by 24 for all n \u03f5 N<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 52n<\/sup> \u2013 1 is divisible by 24<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 52<\/sup> \u2013 1 = 25 \u2013 1 = 24<\/p>\n\n\n\n

P (n) is true for n = 1. Where, P (n) is divisible by 24<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 52k<\/sup> \u2013 1 is divisible by 24<\/p>\n\n\n\n

: 52k<\/sup> \u2013 1 = 24\u03bb \u2026 (i)<\/p>\n\n\n\n

We have to prove,<\/p>\n\n\n\n

52k + 1<\/sup> \u2013 1 is divisible by 24<\/p>\n\n\n\n

52(k + 1)<\/sup> \u2013 1 = 24\u03bc<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= 52(k + 1)<\/sup> \u2013 1<\/p>\n\n\n\n

= 52k<\/sup>.52<\/sup> \u2013 1<\/p>\n\n\n\n

= 25.52k<\/sup> \u2013 1<\/p>\n\n\n\n

= 25.(24\u03bb + 1) \u2013 1 by using equation (1)<\/p>\n\n\n\n

= 25.24\u03bb + 24<\/p>\n\n\n\n

= 24\u03bb<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

20. 32n<\/sup> + 7 is divisible by 8 for all n \u03f5 N<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 32n<\/sup> + 7 is divisible by 8<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 32<\/sup> + 7 = 9 + 7 = 16<\/p>\n\n\n\n

P (n) is true for n = 1. where, P (n) is divisible by 8<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 32k<\/sup> + 7 is divisible by 8<\/p>\n\n\n\n

: 32k<\/sup> + 7 = 8\u03bb<\/p>\n\n\n\n

: 32k<\/sup> = 8\u03bb \u2013 7 \u2026 (i)<\/p>\n\n\n\n

We have to prove,<\/p>\n\n\n\n

32(k + 1)<\/sup> + 7 is divisible by 8<\/p>\n\n\n\n

32k + 2<\/sup> + 7 = 8\u03bc<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= 32(k + 1)<\/sup> + 7<\/p>\n\n\n\n

= 32k<\/sup>.32<\/sup> + 7<\/p>\n\n\n\n

= 9.32k<\/sup> + 7<\/p>\n\n\n\n

= 9.(8\u03bb \u2013 7) + 7 by using equation (i)<\/p>\n\n\n\n

= 72\u03bb \u2013 63 + 7<\/p>\n\n\n\n

= 72\u03bb \u2013 56<\/p>\n\n\n\n

= 8(9\u03bb \u2013 7)<\/p>\n\n\n\n

= 8\u03bc<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

21. 52n + 2<\/sup> \u2013 24n \u2013 25 is divisible by 576 for all n \u03f5 N<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 52n + 2<\/sup> \u2013 24n \u2013 25 is divisible by 576<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 52.1+2<\/sup> \u2013 24.1 \u2013 25<\/p>\n\n\n\n

: 625 \u2013 49<\/p>\n\n\n\n

: 576<\/p>\n\n\n\n

P (n) is true for n = 1. Where, P (n) is divisible by 576<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 52k + 2<\/sup> \u2013 24k \u2013 25 is divisible by 576<\/p>\n\n\n\n

: 52k + 2<\/sup> \u2013 24k \u2013 25 = 576\u03bb \u2026. (i)<\/p>\n\n\n\n

We have to prove,<\/p>\n\n\n\n

52k + 4<\/sup> \u2013 24(k + 1) \u2013 25 is divisible by 576<\/p>\n\n\n\n

5(2k + 2) + 2<\/sup> \u2013 24(k + 1) \u2013 25 = 576\u03bc<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= 5(2k + 2) + 2<\/sup> \u2013 24(k + 1) \u2013 25<\/p>\n\n\n\n

= 5(2k + 2)<\/sup>.52<\/sup> \u2013 24k \u2013 24\u2013 25<\/p>\n\n\n\n

= (576\u03bb + 24k + 25)25 \u2013 24k\u2013 49 by using equation (i)<\/p>\n\n\n\n

= 25. 576\u03bb + 576k + 576<\/p>\n\n\n\n

= 576(25\u03bb + k + 1)<\/p>\n\n\n\n

= 576\u03bc<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

22. 32n + 2<\/sup> \u2013 8n \u2013 9 is divisible by 8 for all n \u03f5 N<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 32n + 2<\/sup> \u2013 8n \u2013 9 is divisible by 8<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 32.1 + 2<\/sup> \u2013 8.1 \u2013 9<\/p>\n\n\n\n

: 81 \u2013 17<\/p>\n\n\n\n

: 64<\/p>\n\n\n\n

P (n) is true for n = 1. Where, P (n) is divisible by 8<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 32k + 2<\/sup> \u2013 8k \u2013 9 is divisible by 8<\/p>\n\n\n\n

: 32k + 2<\/sup> \u2013 8k \u2013 9 = 8\u03bb \u2026 (i)<\/p>\n\n\n\n

We have to prove,<\/p>\n\n\n\n

32k + 4<\/sup> \u2013 8(k + 1) \u2013 9 is divisible by 8<\/p>\n\n\n\n

3(2k + 2) + 2<\/sup> \u2013 8(k + 1) \u2013 9 = 8\u03bc<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= 32(k + 1)<\/sup>.32<\/sup> \u2013 8(k + 1) \u2013 9<\/p>\n\n\n\n

= (8\u03bb + 8k + 9)9 \u2013 8k \u2013 8 \u2013 9<\/p>\n\n\n\n

= 72\u03bb + 72k + 81 \u2013 8k \u2013 17 using equation (1)<\/p>\n\n\n\n

= 72\u03bb + 64k + 64<\/p>\n\n\n\n

= 8(9\u03bb + 8k + 8)<\/p>\n\n\n\n

= 8\u03bc<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

23. (ab) n<\/sup> = an<\/sup> bn<\/sup> for all n \u03f5 N<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): (ab) n<\/sup> = an<\/sup> bn<\/sup><\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): (ab) 1<\/sup> = a1<\/sup> b1<\/sup><\/p>\n\n\n\n

: ab = ab<\/p>\n\n\n\n

P (n) is true for n = 1.<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): (ab) k<\/sup> = ak<\/sup> bk<\/sup> \u2026 (i)<\/p>\n\n\n\n

We have to prove,<\/p>\n\n\n\n

(ab) k + 1 <\/sup>= ak + 1<\/sup>.bk + 1<\/sup><\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= (ab) k + 1<\/sup><\/p>\n\n\n\n

= (ab) k<\/sup> (ab)<\/p>\n\n\n\n

= (ak <\/sup>bk<\/sup>) (ab) using equation (1)<\/p>\n\n\n\n

= (ak + 1<\/sup>) (bk + 1<\/sup>)<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

24. n (n + 1) (n + 5) is a multiple of 3 for all n \u03f5 N.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): n (n + 1) (n + 5) is a multiple of 3<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 1 (1 + 1) (1 + 5)<\/p>\n\n\n\n

: 2 \u00d7 6<\/p>\n\n\n\n

: 12<\/p>\n\n\n\n

P (n) is true for n = 1. Where, P (n) is a multiple of 3<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): k (k + 1) (k + 5) is a multiple of 3<\/p>\n\n\n\n

: k(k + 1) (k + 5) = 3\u03bb \u2026 (i)<\/p>\n\n\n\n

We have to prove,<\/p>\n\n\n\n

(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3<\/p>\n\n\n\n

(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3\u03bc<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= (k + 1) [(k + 1) + 1] [(k + 1) + 5]<\/p>\n\n\n\n

= (k + 1) (k + 2) [(k + 1) + 5]<\/p>\n\n\n\n

= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]<\/p>\n\n\n\n

= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)<\/p>\n\n\n\n

= 3\u03bb + k2<\/sup> + k + 2(k2<\/sup> + 6k + 5) + 2k + 2<\/p>\n\n\n\n

= 3\u03bb + k2<\/sup> + k + 2k2<\/sup> + 12k + 10 + 2k + 2<\/p>\n\n\n\n

= 3\u03bb + 3k2<\/sup> + 15k + 12<\/p>\n\n\n\n

= 3(\u03bb + k2<\/sup> + 5k + 4)<\/p>\n\n\n\n

= 3\u03bc<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

25. 72n<\/sup> + 23n \u2013 3<\/sup>. 3n \u2013 1 is divisible by 25 for all n \u03f5 N<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let P (n): 72n<\/sup> + 23n \u2013 3<\/sup>. 3n \u2013 1 is divisible by 25<\/p>\n\n\n\n

Let us check for n = 1,<\/p>\n\n\n\n

P (1): 72<\/sup> + 20<\/sup>.30<\/sup><\/p>\n\n\n\n

: 49 + 1<\/p>\n\n\n\n

: 50<\/p>\n\n\n\n

P (n) is true for n = 1. Where, P (n) is divisible by 25<\/p>\n\n\n\n

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.<\/p>\n\n\n\n

P (k): 72k<\/sup> + 23k \u2013 3<\/sup>. 3k \u2013 1 is divisible by 25<\/p>\n\n\n\n

: 72k<\/sup> + 23k \u2013 3<\/sup>. 3k \u2013 1<\/sup> = 25\u03bb \u2026 (i)<\/p>\n\n\n\n

We have to prove that:<\/p>\n\n\n\n

72k + 1<\/sup> + 23k<\/sup>. 3k<\/sup> is divisible by 25<\/p>\n\n\n\n

72k + 2<\/sup> + 23k<\/sup>. 3k<\/sup> = 25\u03bc<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

= 72(k + 1)<\/sup> + 23k<\/sup>. 3k<\/sup><\/p>\n\n\n\n

= 72k<\/sup>.71<\/sup> + 23k<\/sup>. 3k<\/sup><\/p>\n\n\n\n

= (25\u03bb \u2013 23k \u2013 3<\/sup>. 3k \u2013 1<\/sup>) 49 + 23k<\/sup>. 3k by using equation (i)<\/p>\n\n\n\n

= 25\u03bb. 49 \u2013 23k<\/sup>\/8. 3k<\/sup>\/3. 49 + 23k<\/sup>. 3k<\/sup><\/p>\n\n\n\n

= 24\u00d725\u00d749\u03bb \u2013 23k <\/sup>. 3k <\/sup>. 49 + 24 . 23k<\/sup>.3k<\/sup><\/p>\n\n\n\n

= 24\u00d725\u00d749\u03bb \u2013 25 . 23k<\/sup>. 3k<\/sup><\/p>\n\n\n\n

= 25(24 . 49\u03bb \u2013 23k<\/sup>. 3k<\/sup>)<\/p>\n\n\n\n

= 25\u03bc<\/p>\n\n\n\n

P (n) is true for n = k + 1<\/p>\n\n\n\n

Hence, P (n) is true for all n \u2208 N.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 12: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 12\u2013Mathematical Induction<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 12\u2013Mathematical Induction PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n