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

1 Answer
Oct 19, 2016

Answer:

See below.

Explanation:

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.