Help needed on a problem regarding diffcult manipulation of binomial coefficients What is the coefficient of $x^{n-1}$ in $$(x+1)^{n}+(x+1)^{n+2}+(x+1)^{n+4}+\dots+(x+1)^{n+2m}\;,$$ where $x,n,m$ are positive integers?  Is there a closed form answer?
 A: By the binomial theorem the coefficient is $\sum_{i=0}^m\binom{n+2i}{n-1}$
A: By the binomial theorem this is
$$\sum_{k=0}^m\binom{n+2k}{n-1}=\sum_{k=0}^m\binom{n+2k}{1+2k}\;.$$
For $n=1$ this is easily seen to be $m+1$, and the case $n=2$ is almost as easy: it’s $2\binom{m+2}2$. 
For $n=3$ it can be expressed as $$\frac16(m+1)(m+2)(4m+9)=4\binom{m+3}3-\binom{m+2}2\;,$$ which already looks rather unpromising. 
For $n=4$ the series is $4+35+126+330+715+\dots$, yielding $4,39,165,495,1210\dots$ as the sequence of values. Since this sequence isn’t in OEIS, I’m not too sanguine about the possibility of coming up with a closed form for the general problem.
A: Hint: Coefficient of $x^k$ in $(1+x)^n$ is $\displaystyle\binom n k$. Further $\displaystyle\binom n k=\displaystyle\binom n {n-k}$
Also, you'll need this $$\binom n r +\binom n {r+1}=\binom {n+1}{r+1}$$

A simple answer would be: $$\sum_{k=0}^m\binom{n+2k}{2k+1}$$ However, a closed form evades me. For consecutive exponents, I have succeeded and I've added it below. 


Here's a full solution for consecutive exponents:
The coefficient of $x^{n-1}$ is $$\begin{align}&\binom {n}{n-1} +\binom{n+1}{n-1} +\binom {n+2}{n-1}+ \cdots +\binom {n+2m}{n-1}\\&=n+1-1+\binom{n+1}{n-1} +\binom {n+2}{n-1}+ \cdots +\binom {n+2m}{n-1}\\&=\binom {n+1}{1}+\binom{n+1}{2} +\binom {n+2}{3}+ \cdots +\binom {n+2m}{2m+1}-1\\&=\binom {n+2m+1}{2m+1} -1\\&=\binom {n+2m+1}{n} -1\end{align}$$ 
