Value of$ \sum^{50}_{k=1}(49+k)! \cdot (50-k)!\cdot (2k-1)$

Question

Finding value of $$\sum^{50}_{k=1}\frac{\binom{50+k}{k}\cdot (2k-1)}{\binom{50}{k}\cdot (50+k)}.$$

Try: Let $$S = \sum^{50}_{k=1}\frac{(50+k)!}{50! \times k!}\times \frac{k! \times (50-k)!}{50!}\cdot \frac{(2k-1)}{(50+k)}$$

$$S = \frac{1}{(50!)^2}\cdot \sum^{50}_{k=1}(49+k)! \cdot (50-k)!\cdot (2k-1)$$

I have seems that it must be in Telescopic series

But could not find How i convert it

Could some help me

Answer 1

You may notice that $(2k-1) = (50+k)-(51-k)$, hence $(50!)^2 S$ can be written as

$$ \sum_{k=1}^{50}(50+k)!(50-k)!-\sum_{k=1}^{50}(49+k)!(51-k)! $$ or, by reindexing the second sum, as $$ \sum_{k=1}^{50}(50+k)!(50-k)!-\sum_{k=0}^{49}(50+k)!(50-k)! $$ with a massive cancellation, leading to $100!-50!^2$ and to the final outcome $\binom{100}{50}-1$.