Prove by induction that an expression is divisible by 11 
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.

I am rather confused by this question. This is my attempt so far:

For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$

The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
 A: hint: You may want to reconsider the way you split the terms at the end.
Note that $64(2^{3k - 1}) + 45(3^k) = 9(2^{3k - 1} + 5(3^k)) + 55(2^{3k - 1})$
A: Keep going!
$64\cdot 2^{3k-1} + 45\cdot 3^k = 9(2^{3k-1} + 5\cdot3^k) + 55\cdot2^{3k-1}$
A: Hint note that: if $k$ is a even number then  also the next number $k+2$ is even 
$$2^{3(k+2)-1}+5\cdot3^{k+2}=2^{3k-1+6}+5\cdot3^{k+2}=64\cdot2^{3k-1}+9\cdot5\cdot3^{k}$$$$=55\cdot2^{3k-1}+9\cdot2^{3k-1}+9\cdot5\cdot3^{k}=55\cdot2^{3k-1}+9\cdot(2^{3k-1}+5\cdot3^{k})$$
A: First, show that this is true for $n=2$:


*

*$\frac{2^{3\cdot2-1}+5\cdot3^1}{11}=7\in\mathbb{N}$


Second, assume that this is true for $n$:


*

*$\frac{2^{3n-1}+5\cdot3^n}{11}=k\in\mathbb{N}$


Third, prove that this is true for $n+2$:


*

*$\frac{2^{3(n+2)-1}+5\cdot3^{n+2}}{11}=\frac{2^{3n+5}+5\cdot3^{n+2}}{11}$

*$\frac{2^{3n+5}+5\cdot3^{n+2}}{11}=\frac{2^6\cdot2^{3n-1}+3^2\cdot5\cdot3^n}{11}$

*$\frac{2^6\cdot2^{3n-1}+3^2\cdot5\cdot3^n}{11}=\frac{64\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}$

*$\frac{64\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}=\frac{55\cdot2^{3n-1}+9\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}$

*$\frac{55\cdot2^{3n-1}+9\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}=\frac{55\cdot2^{3n-1}+9(2^{3n-1}+5\cdot3^n)}{11}$

*$\frac{55\cdot2^{3n-1}+9(2^{3n-1}+5\cdot3^n)}{11}=\frac{55\cdot2^{3n-1}+9\cdot11k}{11}$ assumption used here

*$\frac{55\cdot2^{3n-1}+9\cdot11k}{11}=\frac{11(5\cdot2^{3n-1}+9k)}{11}$

*$\frac{11(5\cdot2^{3n-1}+9k)}{11}=5\cdot2^{3n-1}+9k\in\mathbb{N}$
A: This is the same as proving that $2^{6n-1}+5\cdot3^{2n}$ is divisible by $11$ for all $n$. The case $n=1$ is obvious.
By induction hypothesis, you can assume $2^{6n-1}+5\cdot3^{2n}=11k$, which can be written
$$
2^{6n-1}=11k-5\cdot3^{2n}
$$
Now
\begin{align}
2^{6(n+1)-1}+5\cdot3^{2(n+1)}
&=2^6\cdot2^{6n-1}+45\cdot3^{2n}\\
&=2^6(11k-5\cdot3^{2n})+45\cdot3^{2n}\\
&=11\cdot 2^6k+3^{2n}(45-5\cdot64)
\end{align}
and you're done because $45-5\cdot64=-275=-11\cdot16$.
A: Hint $\ $ Times $2$ it is equivalent to $\,{\rm mod}\ 11\!:\  8^n - 3^n\equiv 0,\,$ i.e.$\ 3^n\equiv (-3)^n\equiv \color{#c00}{(-1)^n} 3^n.\,$ Thus it suffices to prove $\ n$ even $\,\Rightarrow\, \color{#c00}{(-1)^n}\equiv 1,\,$ which is straightforward (by induction or not). 
A: $\begin{align}2^{3}\equiv -3\pmod {11} & \implies 3^n\equiv 2^{3n}\pmod {11} \ \color{blue}{(\text{since}\ n\ \text{is even})}\\& \implies 2^{3n-1}+5\cdot 3^n\equiv 2^{3n-1}+5\cdot 2^{3n}\pmod {11}\\&\implies2^{3n-1}+5\cdot 3^n\equiv 2^{3n-1}(1+5\cdot 2)\pmod {11}\\&\implies 2^{3n-1}+5\cdot 3^n\equiv 0\pmod {11}\end{align}$
A: This might be irrelevant but it is equal to $2^{6n-1}+5 \cdot 9^n=2^{6n-6} \cdot 2^5+5\cdot 9^n$ 
Now since $2^6=64 \equiv -2 (mod 11)$, and $9^n \equiv -2(mod 11)$, the whole expression is equivalent to $(-2)^{n-1} \cdot 32+5 \cdot (-2)^n (mod 11) \equiv (-2)^n \cdot -16+5 \cdot (-2)^n (mod 11) \equiv 11 \cdot (-2)^n (mod 11)$, which is divisible by 11.  
A: Hint. Let $A_k = 2^{3k-1}+5\cdot 3^k = \frac{1}{2}\cdot 8^k+5\cdot 3^k$.
Once you prove that the sequence $\{A_k\}_{k\geq 1}$ fulfills the recurrence relation
$$ A_{k+2} = 11\cdot A_{k+1} -24\cdot A_k $$
you trivially have that $A_{k+2}\equiv -2\cdot A_k\pmod{11}$ holds.
$A_2=77\equiv 0\pmod{11}$ and the conclusion is straightforward.
