Question

If k and n are positive integers and $S_k=1^k+2^k+3^k+.....+n^k,$ then ${\sum }_{r=1}^m$ ${}^{(m+1)}{C}_r{S}_r$ is

• 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

Answer: (A)

$(n+1{)}^{m+1}-(n+1)$

$(1+x{)}^{m+1}=1+{}^{m+1}{C}_1{x}^1+{}^{m+1}{C}_2{x}^2...+{}^{m+1}{C}_{m+1}{x}^{m+1}$
$(1+x{)}^{m+1}-(1+x^{m+1})={}^{m+1}{C}_1{x}^1+{}^{m+1}{C}_2{x}^2...+{}^{m+1}{C}_m{x}^m$
Substituting $x=1,2,3,4$..n in the above equation we get.
$[2^{m+1}-1^{m+1}+(3^{m+1}-2^{m+1})...((n+1{)}^{m+1})-n^{m+1}\left)\right]-n$
$=(n+1{)}^{m+1}-(n)-1$
$=(n+1{)}^{m+1}-(n+1)$
Hence answer is Option A