<s>[INST] Calculate $2 C_{n}^{k-1} + 3 C_{n}^{k-2} + 4 C_{n}^{k-3} + ... + (k + 1)C_{n}^{0}$ Are there any easy ways to calculate this?
$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = 2 C_{2k}^{k-1} + 3 C_{2k}^{k-2} + 4 C_{2k}^{k-3} + ... + (k + 1) C_{2k}^{0} $$
I tried to "turn" the expression:
$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = n \cdot 2^{n-2} - \frac{1}{2} \sum_{i=1}^{k} (2n -i) \cdot C_{2k}^{k-i} $$
But, it doesn't help me.
Thanks for the help!
 [/INST] Consider the formula provided by this answer:

$\sum_{0\leq j\leq n/2}\binom{n}{j}x^j = (x+1)^n - \binom{n}{\left\lfloor\frac{n}{2}\right\rfloor + 1} x^{\left\lfloor\frac{n}{2}\right\rfloor + 1} {_2F_1}\left(1,\left\lfloor\frac{n}{2}\right\rfloor - n + 1;\left\lfloor \frac{n}{2}\right\rfloor + 2;-x\right),$
where $_2F_1$ is the hypergeometric function.

Let $f(x)$ be the function above with $n=2k$, that is
$$f(x) = \sum_{j=0}^{k}\binom{2k}j x^j = (x+1)^{2k} - \binom{2k}{k + 1} x^{k + 1} {_2F_1}\left(1,1-k;k + 2;-x\right).$$
Now write $(x+1)^{2k} = \sum_{j=0}^{2k}\binom{2k}j x^{j}$, so that
$$\begin{align}
(x+1)^{2k} - f(x)
&= \sum_{j=k+1}^{2k}\binom{2k}j x^j
\\&= \sum_{i=1}^{k}\binom{2k}{k+i} x^{k+i}
= \sum_{i=1}^{k}\binom{2k}{k-i} x^{k+i}
\end{align}$$
With $h(x) = (x+1)^{2k} - f(x) = \binom{2k}{k + 1} x^{k + 1} {_2F_1}\left(1,1-k;k + 2;-x\right)$, we hence have
$$x^{1-k}h(x) = \sum_{i=1}^{k}\binom{2k}{k-i} x^{i+1}.$$
Differentiating we get
$$(1-k)x^{-k}h(x) + x^{1-k}h'(x) = \sum_{i=1}^{k}(i+1)\binom{2k}{k-i} x^i,$$
so that your sum can be obtained by evaluating the LHS at $x=1$:
$$(1-k)h(1) + h'(1).$$
This yields a complicated expression in terms of hypergeometric functions:
$$\begin{align}
&&(1-k)&&\binom{2k}{k+1}&&{_2F_1}\left(1,1-k;k + 2;-1\right)
\\+\,\,\,&&(k+1)&&\binom{2k}{k+1}&&{_2F_1}\left(1,1-k;k + 2;-1\right)
\\+\,\,\,&&\frac{k-1}{k+2}&&\binom{2k}{k+1}&&{_2F_1}\left(2,2-k;k + 3;-1\right)
\end{align}
\\= \binom{2k}{k+1}\left(2{_2F_1}\left(1,1-k;k + 2;-1\right) + \frac{k-1}{k+2}{_2F_1}\left(2,2-k;k + 3;-1\right)\right).$$
You can check on WolframAlpha that the formula provided above and the expression in the opening question yield the same results.
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