Question

Prove that 2power n + 6x9power n is always divisible by 7 for any positive integer n.
​

Answer 1

Answer:

I have proved this result by mathematical induction.

Let P(n) denote the statement

$"2^n+6(9^n)$ is divisible by 7"

Put n=1,

P(1):

$2^1+6(9^1)$

$=2+6(9)$

$=2+54$

$=56$ which is divisible by 7

Hence P(1) is true

Assumme that P(k) is true

That is.

$"2^k+6(9^k)$ is divisible by 7"

Then

$2^k+6(9^k)=7m$ where m is an integer................(1)

To prove: P(k+1) is true

That is to prove

$"2^{k+1}+6(9^{k+1})$ is divisible by 7"

Now,

$2^{k+1}+6(9^{k+1})$

$=2^k.2+6(9^k.9)$

$=2^k.2+(6.9^k))9$

$=2^k.2+(7m-2^k)9$

$=2.2^k+63m-9.2^k$

$=63m+2.2^k-9.2^k$

$=63m-7.2^k$

$=7(9m-2^k)$ which is divisible by 7

Therefore, p(k+1) is true

Hence by mathematical induction, P(n) is true for all natural number.