Find the sum of this series:
$\sum_{r=1}^{19}\left( (-1)^{r-1}\frac{r}{{20\choose r}}\right)$
NOTE: ${20\choose r}$ = $\frac{20!}{(20-r)!r!}$
The final answer is 10/11. I do have a way to solve this, but i find that to be wayy too tedious and long. Please don't use concepts that are above high school level.
Thanks in advance!
In general you are dealing with: $$S_n^{(m)} := \sum_{k=0}^{n} (-1)^k k^m \binom{n}{k}^{-1}$$
We really only want to deal with:
$$S_{n}^{(0)}=(1+(-1)^n)\frac{n+1}{n+2}$$ For $n=20$ and also subtract $2=\binom{20}{0}^{-1}+\binom{20}{20}^{-1}$, and negate the expression, i.e:
$$\sum_{r=1}^{19}(-1)^{r-1}\binom{20}{r}^{-1}=2-S_{20}^{(0)}=2-2\frac{21}{22}=\frac{1}{11}$$
Next:
$$\sum_{r=1}^{19} (-1)^{r-1}r\binom{20}{r}^{-1} = 10\sum_{r=1}^{19}(-1)^{r-1}\binom{20}{r}^{-1}=\frac{10}{11}$$
$r$ was replaced by a constant of $10$ due to the $9$ pairs in the sum (Some all positive, the rest all negative): $$r\binom{20}{r}^{-1}+(20-r)\binom{20}{20-r}^{-1}=10\binom{20}{r}^{-1}+10\binom{20}{20-r}^{-1}$$ And the $10$th is of course the central $-10\binom{20}{10}^{-1}$.