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

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

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Answer 1 · 2016-10-19T19:46:00

See below.

Explanation:

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

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

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

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

but if $n > 1$ $n > 1$ we have

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

so inductively the assertion is true.

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

1 Answer

Explanation:

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