Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$: 
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1 
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1 
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1 
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1 
\\
= &16k+4k+4 +1+4r+2+1 
\\
= 
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.

If you want to use induction:
$5^{n+1} + 2\cdot3^n +1 = 5^n\cdot5 + 2\cdot3^{n-1}\cdot3 + 1 = 5^n + 3^{n-1} + 1 + 4\cdot5^n + 4\cdot3^{n-1} = 8K + (4\cdot5^n + 4\cdot3^{n-1})$.
Suffices to show $4\cdot5^n + 4\cdot3^{n-1}$ is divisible by 8.  It's clearly divisible by 4.  So it suffices to show $5^n + 3^{n-1}$ is even.  Which we can do by, heh heh, induction (yes, you can do induction within induction).
$n = 1; 5^1 + 3^0 = 6$ Even.  Induction: $5^{n+1} + 3^n = 5^n + 3^{n-1} + (4\cdot5^n + 2\cdot3^{n-1})$  It's even.
