Writing $(m+2)^n-(m-2)^n$ in summation notation.

Question

I have expanded $(m+2)^n-(m-2)^n$ the following way:

$$(m+2)^n-(m-2)^n = 2 {n \choose 1}m^{n-1}+ \dots + {n \choose n-1}m2^{n-1}-{n \choose n-1}m(-2)^{n-1}+{n \choose n}2^n - {n \choose n}(-2)^n$$

Is my expansion correct and is it possible to write it in compact summation notation?

Answer 1

We have

$$(m+2)^n-(m-2)^n=\sum_{k=0}^n \binom{n}{k}2^km^{n-k}-\sum_{k=0}^n (-1)^k\binom{n}{k}2^km^{n-k}=$$

$$=\sum_{k=0}^n(1-(-1)^k)\binom{n}{k}2^km^{n-k}=\sum_{\substack{k=0\\k\, odd}}^n \binom{n}{k}2^{k+1}m^{n-k}$$

Answer 2

Use the following two versions of the Binomial theorem: $$(x-y)^n=\sum_{k=0}^n (-1)^k\binom{n}{k}x^{n-k}y^k$$ $$(x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$$

Answer 3

Using Newton's binomial theorem

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

we get that

\begin{align} (m+2)^n - (m-2)^n &= \sum_{k=0}^n\,\binom nk\, m^{n-k}\,2^k - \sum_{k=0}^n\,\binom nk\, m^{n-k}\,{(-2)}^k \\&= \sum_{k=0}^n\,\binom nk\, m^{n-k}\,\underbrace{\left(2^k-{(-2)}^k\right)}_{(*)}. \end{align}

When $k$ is even, we have $(*) = 0$. It follows that

$$ (m+2)^n - (m-2)^n = \sum_{j=1}^{\lceil n/2\rceil}\,\binom n{2j-1}\, m^{n-(2j-1)}\,4^j. $$