# Prove by Induction

#### simcan18

##### New member
Can someone with understanding of proof by induction help with this problem?

Prove by induction that 3 raised to 2n+1 + 2 raised to n-1 is divisible by 7 for all numbers greater than/or equal to 1. How do you do the inductive step?

#### simcan18

##### New member
I have done the base case and some of the inductive..which I'm not sure I'm going in the right direction.
Inductive, So does it hold true for n=k+1
3 raised 2(k+1)+1 +2 raised(k+1)-1 = 3 raised 2k+2+1 +2 raised (k+1)-1
= 3 raised 2k+1 x 3 raised2 + 2 raised k x 2 raised 0
=9 x 3 raised 2k+1 + 1 x 2 raised k
= 27 x 3 raised 2k +1x2 raised k

Problem isn't posting correctly

#### Olinguito

##### Well-known member
$3^{2(k+1)+1}+2^{(k+1)-1}$​

$=\quad3^{2k+3}+2^k$

$=\quad9\cdot3^{2k+1}+2\cdot2^{k-1}$

$=\quad7\cdot3^{2k+1}+2\cdot\left(3^{2k+1}+2^{k-1}\right).$

It should be straightforward to proceed from here.

When doing problems of this kind, look at the number you want your expression to be divisible by (in this case $7$) and try and rearrange your expression to involve it.