Question

7^2n + 2^2n – 2 . 3^n – 1 is divisible by 50 by principle of mathematical induction.

(a) True

(b) False

I got this question in an online interview.

This question is from The Principle of Mathematical Induction in section Principle of Mathematical Induction of Mathematics – Class 11

Answer 1

Right choice is (b) False

Easy explanation: P(n) = 7^2n + 2^2n – 2 . 3^n – 1

P(1) = 7^2 + 2^0 . 3^0

P(1) = 50

We now assume that P(k) is divisible by 50.

Therefore, P(k) = 7^2k + 2^2k – 2 . 3^k – 1

To prove P(k + 1) = 7^2(k – 1) + 2^2(k + 1) – 2 . 3^(k + 1) – 1 is divisible by 50

P(k + 1) = 7^2k . 7^2 + 2^2k . 3^k

P(k + 1) = 7^2 ( 7^2k + 2^2n – 2 . 3^k – 1 – 2^2k – 2  . 3^k – 1 ) + 2^2k . 3^k

P(k + 1) = 7^2 x 50c – 7^2 . 2^2k – 2 . 3^k – 1 + 2^2k . 3^k

Since P(k) = 7^2k + 2^2n – 2 . 3^k – 1 – 2^2k – 2 . 3^k – 1 is divisible by 50, it can be written as 50c.

P(k + 1) =  49 x 50C – 49 . 2^2k – 2 . 3^k – 1 + 3 x 4 x 2^2k – 2 . 3^k – 1

P(k + 1) = 49 x 50C – 2^2k – 2 . 3^k – 1 x 37

While the first term is divisible by 50, the second term is not.

Therefore, by principle of mathematical induction, 7^2n + 2^2n – 2 . 3^n – 1 is not divisible by 50.