Demonstrate that $2^n+6^n$ is divisible by $8$ for $n=1,2,3,cdots$ ?

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Answer 1 · 2016-10-19T19:46:00

See below.

For $n=1$ we have $f(1)=8$ which is divisible by $8$
Now, supposing that $f(n)$ is divisible by $8$ then

$f(n) = 2^2+6^2=8 cdot k$

The last step is verify if $f(n+1)$ is divisible by $8$ .

$f(n+1) = 2^(n+1)+6^(n+1) = 2 cdot 2^n+6 cdot 6^n = 6 cdot 2^n + 6 cdot 6^n - 4 cdot 2^n = 6 cdot 8 cdot k-2^2cdot 2^n$

but if $n > 1$ we have

$f(n+1)= 6 cdot 8 cdot k-2^2cdot 2^n$ is divisible by $8$

so inductively the assertion is true.

Demonstrate that 2^n+6^n2^n+6^n is divisible by 88 for n=1,2,3,cdots n=1,2,3,cdots ?