Prove that $3^{2 n} + 7$ is a multiple of $8$ .

Open in App

Solution

Let $p = 3^{2 n} + 7$
$p = {(3^{2})}^{n} + 7 9^{n} - 1 + 8 p = 9^{n} - 1^{n} + 8 a^{n} - b^{n} = (a - b) (a^{n - 1} + a^{n - 2} b + a^{n - 3} b^{2} . . . . . . . . . . . b^{n - 1}) \Rightarrow p = (9 - 1) (9^{n - 1} + 9^{n - 2} .1 + 9^{n - 3} 1^{2} . . . . . . . . . . . 1^{n - 1}) + 8 \Rightarrow p = 8 (9^{n - 1} + 9^{n - 2} .1 + 9^{n - 3} 1^{2} . . . . . . . . . . . 1^{n - 1}) + 8 \Rightarrow p = 8 (9^{n - 1} + 9^{n - 2} .1 + 9^{n - 3} 1^{2} . . . . . . . . . . . 1^{n - 1} + 1)$
which is mutiple of $8$
Hence proved.

Suggest Corrections

Similar questions

n (n + 1) (n + 5)

6

Prove that $5^{k} - 5$ is multiple of 4

3^{2 n + 2} - 8 n - 9

64

1^{2} + 3^{2} + 5^{2} + . . . + {(2 n - 1)}^{2} = \frac{n}{3} (2 n - 1) (2 n + 1)