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!
$$S=2 {n \choose {k-1}}+3{n \choose {k-2}}+4{n \choose {k-3}}+....+k{n \choose 1}+(k+1){n \choose 0}$$
Now writing Sum $S$ in reverse order,
$$S=(k+1){n \choose 0}+k{n \choose 1}+....+3{n \choose {k-2}}+2 {n \choose {k-1}}$$
$$\implies S=k \bigg[{n \choose 0}+{n \choose 1}+....+{n \choose {k-2}}+{n \choose {k-1}}\bigg]+{n \choose 0}-\bigg[1{n \choose 2}+2{n \choose 3}+....+(k-3){n \choose {k-2}}+(k-2){n \choose {k-1}}\bigg]$$
Let $$P={n \choose 0}+{n \choose 1}+....+{n \choose {k-2}}+{n \choose {k-1}}$$
$$Q=1{n \choose 2}+2{n \choose 3}+....+(k-3){n \choose {k-2}}+(k-2){n \choose {k-1}}$$
As $$(1+x)^{n}={n \choose 0}+{n \choose 1}x+{n \choose 2}x^2+....+ {n \choose {k-1}}x^{k-1}+{n \choose k}x^{k}+....+{n \choose n}x^{n-1}$$...$(1)$
On putting $x=1$ we get,
$$P=2^{n}-\sum_{r=0}^{n-k}{n \choose k+r}$$
To get Sum $Q$ We will First Divide Equation $1$ by $x$ then differentiate w.r.t $x$ and then Put $x=1$ we have $Q$ as
$$Q=\Big[n(2)^{n-1}-2^{n}\Big]+1-\sum_{r=0}^{n-k}{n \choose k+r}(k+r-1)$$
So our Sum $S$ Becomes
$$S=k \bigg[2^{n}-\sum_{r=0}^{n-k}{n \choose k+r}\bigg]+1-\Bigg[\Big[n(2)^{n-1}-2^{n}\Big]+1-\sum_{r=0}^{n-k}{n \choose k+r}(k+r-1)\Bigg]$$
On simplifying $S$ we have,
$$S=2^{n}\Big[k+1-\frac{n}{2}\Big]+\sum_{r=0}^{n-k}{n \choose k+r}(r-1)$$
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.
