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Question: If k and n are positive integers and \({{S}_{k}}={{1}^{k}}+{{2}^{k}}+\cdots +{{n}^{k}}\), then \(\su...

If k and n are positive integers and Sk=1k+2k+⋯+nk{{S}_{k}}={{1}^{k}}+{{2}^{k}}+\cdots +{{n}^{k}}, then ∑r=1mmCrSr\sum\limits_{r=1}^{m}{^{m}{{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 \begin{aligned} & \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{aligned}

Explanation

Solution

Hint: Use the fact that the expansion of binomial (1+x)n{{\left( 1+x \right)}^{n}} is given by (1+x)n=1+nC1x+nC2x2+⋯+nCnxn{{\left( 1+x \right)}^{n}}=1{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+\cdots {{+}^{n}}{{C}_{n}}{{x}^{n}}. Hence prove that (1+x)n−xn−1=∑r=1n−1nCrxr{{\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 ∑r=1m+1m+1CrSr\sum\limits_{r=1}^{m+1}{^{m+1}{{C}_{r}}{{S}_{r}}} in the form of Telescopic series, i.e. series where an=f(n)−f(n−1){{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 Sk=1k+2k+⋯+nk{{S}_{k}}={{1}^{k}}+{{2}^{k}}+\cdots +{{n}^{k}}
Now, we know that (1+x)m+1=∑r=0m+1m+1Crxr{{\left( 1+x \right)}^{m+1}}=\sum\limits_{r=0}^{m+1}{^{m+1}{{C}_{r}}{{x}^{r}}}
Put x =1, we get
(1+1)m+1=∑r=0m+1m+1Cr1r=1+∑r=1mm+1Cr1r+1m+1{{\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
(1+2)m+1=∑r=0m+1m+1Cr2r=1+∑r=1mm+1Cr2r+2m+1 ⋮           =    ⋮ \begin{aligned} & {{\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{aligned}
Put x = n, we get
(1+n)m+1=∑r=0m+1m+1Crnr=1+∑r=1mm+1Crnr+nm+1{{\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
2m+1+3m+1+⋯+(n+1)m+1=(1+∑r=1mm+1Cr1r+1m+1)+(1+∑r=1mm+1Cr2r+2m+1)+(1+∑r=1mm+1Cr2r+3m+1)+⋯+(1+∑r=1mm+1Crnr+nm+1){{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 1m+1,2m+1,⋯ ,nm+1{{1}^{m+1}},{{2}^{m+1}},\cdots ,{{n}^{m+1}} to LHS, we get
−1m+1+2m+1−2m+1+⋯+nm+1−nm+1+(n+1)m+1=(1+1+⋯+1)+∑r=1mnCr(1r+2r+⋯+nr)-{{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+∑r=1mm+1CrSr=(n+1)m+1−1n+\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
∑r=1mm+1CrSr=(n+1)m+1−(n+1)\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
∑r=1mm+1CrSr=2C1S1=2(1+2+⋯+n)=n(n+1)\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
(n+1)m+1−(n+1)=(n+1)2−(n+1)=n(n+1){{\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

&\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{aligned}$$ 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.