Thank you for reporting, we will resolve it shortly
Q.
If $k$ be positive integer and $S_{k}=1 k+2 k+\ldots+n^{k}$, then find
the value of $\displaystyle\sum_{r=1}^{m}{ }^{m+1} C_{r} s_{r}$ if $n=10, m =11$
Binomial Theorem
Solution:
We have,$ \displaystyle \sum_{r=1}^{m}{ }^{m+1} C_{r} s_{r}=\displaystyle\sum_{r=1}^{m}{ }^{m+1} C_{r}\left(1^{r}+2^{r}+\ldots+n^{r}\right)$
$=\displaystyle\sum_{k=1}^{n}\left\{\displaystyle\sum_{r=1}^{m}{ }^{m+1} C_{r} k^{r}\right\}$
$=\displaystyle\sum_{k=1}^{n}\left[\left\{\displaystyle\sum_{r=0}^{m+1}{ }^{m+1} C_{r} k^{r}\right\}-{ }^{m+1} C_{0}-{ }^{m+1} C_{m+1} k^{m+1}\right]$
$=\displaystyle\sum_{k=1}^{n}\left\{(1+k)^{m+1}-1-k^{m+1}\right\}=\displaystyle\sum_{k=1}^{n}\left\{(1+k)^{m+1}-k^{m+1}\right\}-\displaystyle\sum_{k=1}^{n} 1$
$=\displaystyle\sum_{k=1}^{n}\left\{(1+k)^{m+1}-k^{m+1}\right\}-n$
$=\left\{\left(2^{m+1}-1^{m+1}\right)+\left(3^{m+1}-2^{m+1}\right)+\ldots\right.$
$\left.+\left\{(n+1)^{m+1}-n^{m+1}\right\}\right\}-n$
$=\left\{(n+1)^{m+1}-1\right\}-n=(n+1)^{m+1}-(n+1)$
Now put $n=10$ and $m=11$ then
$(n+1)^{m+1}-(n+1)=11^{12}-11$