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Calculate:- $$\sum_{r=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}$$ And generalise the result if possible.

I've tried to reduce this to a telescopic sum but could not do it.

I've also made a recurrence relation:-

If S$_{n}$ = $\sum_{r=1}^{n-1} \frac{(-1)^{r-1}r}{n \choose r}$, then S$_n$ = $\frac{n+1}{2n}$$S_{n-1}$ + $\frac{{(-1)}^n}{n}$, but could not go further.

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    $\begingroup$ Please show what you've tried so far. $\endgroup$
    – David Krell
    Commented Mar 21 at 10:21
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    $\begingroup$ Сlarify what the summation index $i$ or $r$ should be? $\endgroup$
    – Leox
    Commented Mar 21 at 11:10
  • $\begingroup$ $$\sum _{r=1}^{n-1} \frac{(-1)^{r-1} r}{\binom{n}{r}}=\frac{1+n+(-1)^n (-1+n (2+n))}{(2+n) (3+n)}$$ where: $n=2024$. $\endgroup$
    – Mariusz Iwaniuk
    Commented Mar 21 at 15:25
  • $\begingroup$ Can you please show the derivation? $\endgroup$
    – Kutta Khan
    Commented Mar 27 at 13:55

1 Answer 1

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$$\sum_{i=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}= \frac{(-1)^{r-1}r}{2024 \choose r}\sum_{i=1}^{2023} 1=\frac{(-1)^{r-1}r \,2023}{2024 \choose r}.$$

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