Single summation form

Question

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?

Answer 1

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*}

Answer 2

$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} = $$

$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} - \binom{n+1}{n+1}\pi^0(\cot \alpha)^{n+1} = $$

$$\sum_{k=0}^{n} \left[\binom{n}{k} \pi^{n-k} - \binom{n+1}{k} \pi^{n+1-k}\right ](\cot\alpha)^{k} - (\cot \alpha)^{n+1} = $$

$$ - (\cot \alpha)^{n+1} + \sum_{k=0}^{n} \left[\binom{n}{k} - \pi\binom{n+1}{k}\right ]\pi^{n-k}(\cot\alpha)^{k}$$

Or, arguably better as Arun Badajena notices:

$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} = $$

$$(\pi + \cot\alpha)^n - (\pi + \cot\alpha)^{n+1} =$$ $$(1 - \pi - \cot\alpha)(\pi + \cot\alpha)^n = $$ $$(1 - \pi - \cot\alpha)\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k} $$

Answer 3

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.

Answer 4

You can just split off the last term of the second sum $$\begin{align*} \sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} \\ = \sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}-(\cot \alpha)^{n+1}\\ =\sum_{k=0}^{n}\left[\binom{n}{k}+\pi\binom{n+1}k\right] \pi^{n-k}(\cot\alpha)^{k}-(\cot \alpha)^{n+1} \end{align*}$$