Single summation form I want to write as a sigle summation of
\begin{align*}
\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}.
\end{align*}
How to change upper limit of any one of the summations?
 A: $$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} = $$
$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} - \binom{n+1}{n+1}\pi^0(\cot \alpha)^{n+1} = $$
$$\sum_{k=0}^{n} \left[\binom{n}{k} \pi^{n-k} - \binom{n+1}{k} \pi^{n+1-k}\right ](\cot\alpha)^{k} - (\cot \alpha)^{n+1} = $$
$$ - (\cot \alpha)^{n+1} + \sum_{k=0}^{n} \left[\binom{n}{k}  - \pi\binom{n+1}{k}\right ]\pi^{n-k}(\cot\alpha)^{k}$$
Or, arguably better as Arun Badajena notices:
$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} = $$
$$(\pi + \cot\alpha)^n - (\pi + \cot\alpha)^{n+1} =$$
$$(1 - \pi - \cot\alpha)(\pi + \cot\alpha)^n = $$
$$(1 - \pi - \cot\alpha)\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k} $$
A: You can use Binomial theorem, then given expression will be:
$$(\pi+\cot \alpha)^n-(\pi+\cot \alpha)^{n+1}=(1-\pi-\cot \alpha)(\pi+\cot \alpha)^n$$
Now, use the Binomial theorem to get a single summation.
A: Second term, evaluated up to $n$ in the sum and then placing out the $k=n+1$ term.
\begin{align*}
\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} &= \sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}  +\cot^{n+1} \alpha
\end{align*}
A: You can just split off the last term of the second sum
$$\begin{align*}
\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}
\\
=
\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}-(\cot \alpha)^{n+1}\\
=\sum_{k=0}^{n}\left[\binom{n}{k}+\pi\binom{n+1}k\right] \pi^{n-k}(\cot\alpha)^{k}-(\cot \alpha)^{n+1}
\end{align*}$$
