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7^2n + 2^2n – 2 . 3^n – 1 is divisible by 50 by principle of mathematical induction.(a) True(b) FalseI 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»

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.


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