Question

If $k$ be positive integer and $S_k=1k+2k+\dots +n^k$, then find the value of $\sum _{r=1}^m{}^{m+1}{C}_r{s}_r$ if $n=10,m=11$

• A. $11^{12}-11$
• B. $10^{11}-10$
• C. $10^{12}-11$
• D. None of these

Answer 1

Answer: (A)

We have,$\sum _{r=1}^m{}^{m+1}{C}_r{s}_r=\sum _{r=1}^m{}^{m+1}{C}_r\left(1^r+2^r+\dots +n^r\right)$
$=\sum _{k=1}^n\left\{\sum _{r=1}^m{}^{m+1}{C}_r{k}^r\right\}$
$=\sum _{k=1}^n\left[\left\{\sum _{r=0}^{m+1}{}^{m+1}{C}_r{k}^r\right\}-{}^{m+1}{C}_0-{}^{m+1}{C}_{m+1}{k}^{m+1}\right]$
$=\sum _{k=1}^n\left\{(1+k{)}^{m+1}-1-k^{m+1}\right\}=\sum _{k=1}^n\left\{(1+k{)}^{m+1}-k^{m+1}\right\}-\sum _{k=1}^{n}1$
$=\sum _{k=1}^n\left\{(1+k{)}^{m+1}-k^{m+1}\right\}-n$
$=\{\left(2^{m+1}-1^{m+1}\right)+\left(3^{m+1}-2^{m+1}\right)+\dots$
$+\left\{(n+1{)}^{m+1}-n^{m+1}\right\}\}-n$
$=\left\{(n+1{)}^{m+1}-1\right\}-n=(n+1{)}^{m+1}-(n+1)$
Now put $n=10$ and $m=11$ then
$(n+1{)}^{m+1}-(n+1)=11^{12}-11$