72n + 22n – 2. 3n – 1 is divisible by 50 by principle of mathematical induction.

72n + 22n – 2. 3n – 1 is divisible by 50 by principle of mathematical induction. Correct Answer False

P(n) = 72n + 22n – 2 . 3n – 1 P(1) = 72 + 20 . 30 P(1) = 50 We now assume that P(k) is divisible by 50. Therefore, P(k) = 72k + 22k – 2 . 3k – 1 To prove P(k + 1) = 72(k – 1) + 22(k + 1) – 2 . 3(k + 1) – 1 is divisible by 50 P(k + 1) = 72k . 72 + 22k . 3k P(k + 1) = 72 ( 72k + 22n – 2 . 3k – 1 – 22k – 2 . 3k – 1 ) + 22k . 3k P(k + 1) = 72 x 50c – 72 . 22k – 2 . 3k – 1 + 22k . 3k Since P(k) = 72k + 22n – 2 . 3k – 1 – 22k – 2 . 3k – 1 is divisible by 50, it can be written as 50c. P(k + 1) = 49 x 50C – 49 . 22k – 2 . 3k – 1 + 3 x 4 x 22k – 2 . 3k – 1 P(k + 1) = 49 x 50C – 22k – 2 . 3k – 1 x 37 While the first term is divisible by 50, the second term is not. Therefore, by principle of mathematical induction, 72n + 22n – 2 . 3n – 1 is not divisible by 50.