Question

prove that 2 ^ n + 6×9^n is always divisible by 7 for any positive integer n​

Answer 1

Answer:

put n=1,2,3 every term will be divided by 7

having remainder 0

Answer 2

Answer and Explanation :

Given : Expression $2^n+6\times 9^n$

To prove that expression is always divisible by 7 for any positive integer n?

Solution :

Using Principal of mathematical induction PMI,

Put n=1,

$2^n+6\times 9^n$

$2^1+6\times 9^1$

$2+54$

$56$

56 is divisible by 7.

So, For n=1 it is true.

Now, Assume that for n=k it is true.

So, $2^k+6\times 9^k$ is divisible by 7.

i.e. $2^k+6\times 9^k=7m$ where m is an integer.

To prove for n=k+1

Put n=k+1

$=2^{k+1}+6\times 9^{k+1}\\=2^k.2+6(9^k.9)\\=2^k.2+(6.9^k))9\\=2^k.2+(7m-2^k)9\\=2.2^k+63m-9.2^k\\=63m+2.2^k-9.2^k\\=63m-7.2^k\\=7(9m-2^k)$

Which is divisible by 7.

So, For n=k+1 it is true.

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