Question

If $k$ and $n$ be the two positive integers such that $S_t=1^t+2^t+\cdots +n^t$, then $\sum _{r=1}^m{(}^{m+1}{C}_r{S}_r)$ is equal to:

• A. $(n+1{)}^{m+1}-(n+1)$
• B. $(n+1{)}^{m+1}+(n+1)$
• C. $(n-1{)}^{m+1}-(n-1)$
• D. None of these

Answer 1

The correct option is A $(n+1{)}^{m+1}-(n+1)$

We have,
$\sum _{r=1}^m{(}^{m+1}{C}_r{S}_r)$
$=\sum _{r=1}^m{}^{m+1}{C}_r(1^r+2^r+\cdots +n^r)$

$=\sum _{k=1}^n[\left(\sum _{r=0}^{m+1}{}^{m+1}{C}_r{k}^r\right)-{}^{m+1}{C}_0{-}^{m+1}{C}_{m+1}{k}^{m+1}]$
$=\sum _{k=1}^n\left(\right(1+k{)}^{m+1}-1-k^{m+1})$
$=\sum _{k=1}^n\left(\right(1+k{)}^{m+1}-k^{m+1})-\sum _{k=1}^{n}1$
$=\sum _{k=1}^n\left(\right(1+k{)}^{m+1}-k^{m+1})-n$
$=\left[\right(2^{m+1}-1^{m+1})+(3^{m+1}-2^{m+1})+\cdots +((n+1{)}^{m+1}-n^{m+1})]-n$
$=\left[\right(n+1{)}^{m+1}-1)-n]$
$=(n+1{)}^{m+1}-(n+1)$