How do I rewrite this expression so that it can be used with the binomial theorem?

Question

I've been stuck on this problem on my homework for a while. I have a expression

$$\sum_{k=0}^n {n \choose k}5^{3n+k}(-6)^{2k-2}$$

which I need to re-write in the form of

$$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k$$

so that I can use the binomial theorem and turn it into closed form.

I'm able to figure out the $(-6)^{2k-2} = ((-6)^2)^{k-1} = 36^{k-1} = \frac{36^k}{36}$ part of the expression, but I can't figure out how to rewrite $5^{3n+k}$ as $x^{n-k}$.

Answer 1

Hint: $5^{3n+k}=5^{3n}5^k = (5^3)^n(5^{-1})^{-k}$... Do you see how to continue?

Hint 2: $(5^3)^n(5^{-1})^{-k} = 125^n5^n(5^{-1})^n(5^{-1})^{-k}$...

Answer 2

Here's an alternative approach you might want to consider:

$$\sum_{k=0}^n {n \choose k}5^{3n+k}(-6)^{2k-2} = \dfrac{5^{3n}}{36}\sum_{k=0}^n {n \choose k}5^{k}6^{2k} = \dfrac{5^{3n}}{36}\sum_{k=0}^n {n \choose k} 180^k = \dfrac{5^{3n}}{36}(180+1)^n = \dfrac{5^{3n}}{36}181^n = \dfrac{22625^{n}}{36}$$

Answer 3

Hint: Another variant to treat $5^{3n+k}$ is to get $n-k$ via \begin{align*} n=(n-k)+k \end{align*}

We obtain \begin{align*} 5^{3n+k}=5^{3(n-k)+3k+k}=5^{3(n-k)+4k}=\left(5^3\right)^{n-k}\left(5^4\right)^k \end{align*}

Since $5^3=125$ and $5^4=625$ we obtain together with $(-6)^{2k-2}=\frac{1}{36}\cdot36^k$

\begin{align*} \frac{1}{36}\sum_{k=0}^n\binom{n}{k}125^{n-k}(625\cdot 36)^k=\ldots \end{align*}