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How can I prove that following is a multiple of 11 with the induction method? $$3^{3n+2}+2^{4n+1}$$

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    Meta-step: try. {Step #1: show it for $n=0$. Step #2: do the induction step -- if Step #2 is not successful, show us where you are stuck.} – 2017-01-30
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    Do you know congruence or modular arithmetic such as. $\,16\equiv 5\pmod{\!11}\,?\ \ $ – 2017-01-30

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Note that any numbers of the form $C_n=Aa^n+Bb^n$ satisfy the recurrence $C_{n+2}=(a+b)C_{n+1}-abC_n$. From this it follows that any number which is a factor of $C_n$ and $C_{n+1}$ will divide all the $C_r$ for $r\ge n$. (one can reach conclusions about lower values of $r$ as well with a little care)

Here we have $C_n=9\cdot 27^n+2\cdot 16^n$ and $C_0=11, C_1=275$.

I mention it not because it is the easiest solution, but because it explains why this persistence of divisibility kind of problem is so prevalent, and makes it easy to construct a range of examples, which are normally practice for applying induction.

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Compute modulo 11 $$3^{3n+2}+2^{4n+1} \equiv 9\times 27^n + 2\times 16^n \equiv 9\times 5^n+2\times 5^n \equiv 11\times 5^n \equiv 0 \pmod {11}$$

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Using induction: Test for base case $n=0$: $$3^2+2^1=11$$ $11$ is a multiple of $11$. Hence, true for $n=0$.

Inductive step: Assume true for $n=k$: $$\frac{3^{3k+2}+2^{4k+1}}{11}=p$$ Where $p \in \mathbb{N}$. $$3^{3k+2}+2^{4k+1}=11p \tag{1}$$ For $n=k+1$: $$3^{3k+5}+2^{4k+5}\tag{2}$$ Now, we will try to put $(2)$ in terms of $p$ using equation $(1)$. $$27\cdot 3^{3k+2}+16\cdot 2^{4k+1}=\color{blue}{16\cdot 3^{3k+2}}+11\cdot 3^{3k+2}+\color{blue}{16\cdot 2^{4k+1}}$$ The $\color{blue}{\text{blue}}$ text can be put in terms of $p$: $$\color{blue}{16\cdot 11p}+11\cdot 3^{3k+2}$$ Now, note that each of these terms is divisible by $11$, since $k\in \mathbb{N}$ and $p\in \mathbb{N}$.

Therefore $n=k+1$ follows from $n=k$. Thus, the statement is true for all $n\in \mathbb{N}$.

If you have any doubts or questions, please do not hesitate to ask.

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    Is the base case enough to infer — $\frac{3^{3k+2}+2^{4k+1}}{11}=p$. and $ p∈N$? What if the case of 0 is one of the few cases that this holds true? I'm sorry if this question isn't logical. – 2017-01-30
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    Just the base case itself won't prove it, that only shows that it is true for $n=0$. You must do the inductive step to show that $n=k+1$ follows from $n=k$, which I've done to show that it is true for all $n\in \mathbb{N}$ – 2017-01-30
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    @Niki $$\begin{align}{\rm i.e.}\quad f(n) &= 9(27^n) + 2(16^n)\\ {\rm and}\qquad \color{#c00}{27} &= \color{#c00}{16} + 11\\ \Rightarrow\ \ f(n\!+\!1) &= 16\, f(n) + 11(\cdots)\\ {\rm therefore}\quad 11\mid f(n\!+\!1)\ \, &{\rm if}\,\ 11\mid f(n) \end{align}$$ $\ \begin{align}\text{Said more simply:}\quad {\rm mod}\ 11\!:\,\ \color{#c00}{27\equiv 16}\Rightarrow &\ 9(\color{#c00}{27}^n)+2(16^n)\\ \equiv &\ 9(\color{#c00}{16}^n)+2(16^n)\\ \equiv &\ 11(16^n)\\ \equiv&\ 0\end{align}\qquad $ – 2017-01-30
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Base case, $n=0$: $\quad 3^{0+2}+2^{0+1} = 9+2 = 11 \quad \checkmark$

Induction hypothesis: $\quad 3^{3n+2}+2^{4n+1} = 11k$

$\begin{align} 3^{3(n+1)+2}+2^{4(n+1)+1} &= 27\cdot3^{3n+2}+16\cdot 2^{4n+1} \\ &= 22\cdot3^{3n+2}+5\cdot3^{3n+2}+11\cdot 2^{4n+1}+5\cdot 2^{4n+1}\\ &= 22\cdot3^{3n+2}+11\cdot 2^{4n+1}+5(3^{3n+2}+ 2^{4n+1})\\ &= 11(2\cdot3^{3n+2}+ 2^{4n+1})+5(11k)\\ &= 11(2\cdot3^{3n+2}+ 2^{4n+1}+5k)\quad \text{as required}\\ \end{align}$

(without induction):

$3^3 \equiv 27 \equiv 5 \bmod 11$ and $2^4 \equiv 16 \equiv 5\bmod 11$

So $3^{3n+2}+2^{4n+1} \equiv 9\cdot 5^n+2\cdot 5^n \equiv 11\cdot 5^n \equiv 0 \bmod 11$