${S}_{n}=\sum _{r=2}^{n}\frac{(-1{)}^{r}{\textstyle (}\genfrac{}{}{0ex}{}{n}{r}{\textstyle )}}{r-1}$

auto23652im
2022-07-12
Answered

Consider the sum

${S}_{n}=\sum _{r=2}^{n}\frac{(-1{)}^{r}{\textstyle (}\genfrac{}{}{0ex}{}{n}{r}{\textstyle )}}{r-1}$

${S}_{n}=\sum _{r=2}^{n}\frac{(-1{)}^{r}{\textstyle (}\genfrac{}{}{0ex}{}{n}{r}{\textstyle )}}{r-1}$

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