Question

If k and n are positive integers and ${{S}_{k}}={{1}^{k}}+{{2}^{k}}+\cdots +{{n}^{k}}$, then $\sum\limits_{r=1}^{m}{^{m}{{C}_{r}}{{S}_{r}}}$ is equal to
$\begin{align}
  & \left[ a \right]\ {{\left( n+1 \right)}^{m+1}}-\left( n+1 \right) \\
 & [b]\ {{\left( n+1 \right)}^{m+1}}+\left( n+1 \right) \\
 & [c]\ {{\left( n-1 \right)}^{m+1}}-\left( n-1 \right) \\
 & [d]\ \text{None of these} \\
\end{align}$

Answer 1

Hint: Use the fact that the expansion of binomial ${{\left( 1+x \right)}^{n}}$ is given by ${{\left( 1+x \right)}^{n}}=1{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+\cdots {{+}^{n}}{{C}_{n}}{{x}^{n}}$. Hence prove that ${{\left( 1+x \right)}^{n}}-{{x}^{n}}-1=\sum\limits_{r=1}^{n-1}{^{n}{{C}_{r}}{{x}^{r}}}$. Replace n by m+1, and put successively x = 1, 2, …,n and add the resulting expressions. Hence express the sum $\sum\limits_{r=1}^{m+1}{^{m+1}{{C}_{r}}{{S}_{r}}}$ in the form of Telescopic series, i.e. series where ${{a}_{n}}=f\left( n \right)-f\left( n-1 \right)$. In a telescopic series, alternate terms cancel out leaving the first and the last term only. Hence find the sum and verify which of the options are correct.

Complete step-by-step solution -
We have ${{S}_{k}}={{1}^{k}}+{{2}^{k}}+\cdots +{{n}^{k}}$
Now, we know that ${{\left( 1+x \right)}^{m+1}}=\sum\limits_{r=0}^{m+1}{^{m+1}{{C}_{r}}{{x}^{r}}}$
Put x =1, we get
${{\left( 1+1 \right)}^{m+1}}=\sum\limits_{r=0}^{m+1}{^{m+1}{{C}_{r}}{{1}^{r}}=1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{1}^{r}}}+{{1}^{m+1}}}$
Put x = 2, we get
$\begin{align}
  & {{\left( 1+2 \right)}^{m+1}}=\sum\limits_{r=0}^{m+1}{^{m+1}{{C}_{r}}{{2}^{r}}=1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{2}^{r}}}+{{2}^{m+1}}} \\
 & \vdots \ \ \ \ \ \ \ \ \ \ \ =\ \ \ \ \vdots \\
\end{align}$
Put x = n, we get
${{\left( 1+n \right)}^{m+1}}=\sum\limits_{r=0}^{m+1}{^{m+1}{{C}_{r}}{{n}^{r}}=1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{n}^{r}}}+{{n}^{m+1}}}$
Adding these equation, we get
${{2}^{m+1}}+{{3}^{m+1}}+\cdots +{{\left( n+1 \right)}^{m+1}}=\left( 1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{1}^{r}}+{{1}^{m+1}}} \right)+\left( 1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{2}^{r}}+{{2}^{m+1}}} \right)+\left( 1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{2}^{r}}+{{3}^{m+1}}} \right)+\cdots +\left( 1+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{n}^{r}}+{{n}^{m+1}}} \right)$
Transposing the terms ${{1}^{m+1}},{{2}^{m+1}},\cdots ,{{n}^{m+1}}$ to LHS, we get
$-{{1}^{m+1}}+{{2}^{m+1}}-{{2}^{m+1}}+\cdots +{{n}^{m+1}}-{{n}^{m+1}}+{{\left( n+1 \right)}^{m+1}}=\left( 1+1+\cdots +1 \right)+\sum\limits_{r=1}^{m}{^{n}{{C}_{r}}\left( {{1}^{r}}+{{2}^{r}}+\cdots +{{n}^{r}} \right)}$
On LHS alternate terms cancel each other except the first and the last term.
Hence, we have
$n+\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{S}_{r}}={{\left( n+1 \right)}^{m+1}}-1}$
Subtracting n from both sides, we get
$\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{S}_{r}}}={{\left( n+1 \right)}^{m+1}}-\left( n+1 \right)$
Hence option [a] is correct.

Note: We can verify the correctness of our solution by checking the correctness of our result of m = 1, 2.
For m = 1, we have
$\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{S}_{r}}}{{=}^{2}}{{C}_{1}}{{S}_{1}}=2\left( 1+2+\cdots +n \right)=n\left( n+1 \right)$
Also, we have
${{\left( n+1 \right)}^{m+1}}-\left( n+1 \right)={{\left( n+1 \right)}^{2}}-\left( n+1 \right)=n\left( n+1 \right)$
LHS = RHS for m = 1.
For m = 2, we have
\[\begin{align}
&\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{S}_{r}}}{{=}^{3}}{{C}_{1}}{{S}_{1}}{{+}^{3}}{{C}_{2}}{{S}_{2}}=3\left( 1+2+\cdots +n \right)+3\left( {{1}^{2}}+{{2}^{2}}+\cdots +{{n}^{2}} \right) \\
 & =3\left( \dfrac{n\left( n+1 \right)}{2}+\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)=\dfrac{3\left( n \right)\left( n+1 \right)}{2}\left( 1+\dfrac{2n+3}{3} \right)=n\left( n+1 \right)\left( n+2 \right) \\
\end{align}\]
Also, we have
${{\left( n+1 \right)}^{m+1}}-\left( n+1 \right)={{\left( n+1 \right)}^{3}}-\left( n+1 \right)=\left( n+1 \right)\left( {{n}^{2}}+2n+1-1 \right)=n\left( n+1 \right)\left( n+2 \right)$
Hence, LHS = RHS for m = 2.
Hence our solution is verified to be correct.