Question

If k,n are positive integers and ${{s}_{k}}={{1}^{k}}+{{2}^{k}}+...+{{n}^{k}}$ , then show that $\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}={{\left( n+1 \right)}^{m+1}}-\left( n+1 \right)}$ and evaluate ${{s}_{4}}.$ $$$$

Answer 1

We expand $\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}$ using ${{s}_{k}}$ and then collect the terms with same base so that we can use the binomial expansion of ${{\left( 1+x \right)}^{a}}$ where $a=1,2,3,...n$. We then simplify to get the proof. We use the expression of sum of first$n$ which is ${{s}_{1}}$, squared $n$ terms which is ${{s}_{2}}$ and cubed $n$ terms which is ${{s}_{3}}$ and the value $n=4$ in proof statement to get ${{s}_{4}}.$$$$$

Complete step-by-step solution:
We know that the binomial expansion of ${{\left( 1+x \right)}^{a}}$ is,

& {{\left( 1+x \right)}^{a+1}}{{=}^{a+1}}{{C}_{0}}{{+}^{a+1}}{{C}_{1}}x{{+}^{a+1}}{{C}_{1}}{{x}^{2}}...{{+}^{a+1}}{{C}_{a+1}}{{x}^{a+1}} \\\ & \Rightarrow {{\left( 1+x \right)}^{a+1}}-1-{{x}^{a+1}}{{=}^{a+1}}{{C}_{1}}x{{+}^{a+1}}{{C}_{1}}{{x}^{2}}...{{+}^{a+1}}{{C}_{a}}{{x}^{a}}....(1) \\\ \end{aligned}$$ We are given in the question that $${{s}_{k}}={{1}^{k}}+{{2}^{k}}+...+{{n}^{k}}$$ We are also given to prove, $$\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}={{\left( n+1 \right)}^{m+1}}-\left( n+1 \right)}$$ Let us expand the left hand side of the above summation and have $$\sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}=}\left( ^{m+1}{{C}_{1}} \right){{s}_{1}}+\left( ^{m+1}{{C}_{2}} \right){{s}_{2}}+...+\left( ^{m+1}{{C}_{m}} \right){{s}_{m}}$$ We substitute the value of ${{s}_{k}}$ for each value of $k=1,2,..,m$ and have, $$\begin{aligned} & \sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}{{=}^{m+1}}{{C}_{1}}\left( {{1}^{1}}+{{2}^{1}}+...+{{n}^{1}} \right){{+}^{m+1}}{{C}_{2}}\left( {{1}^{2}}+{{2}^{2}}+...+{{n}^{2}} \right)+ \\\ & ...{{+}^{m+1}}{{C}_{m}}\left( {{1}^{m}}+{{2}^{m}}+...+{{n}^{m}} \right) \\\ \end{aligned}$$ Let us collect the terms with base 1 and who have exponents $1,2,..m$. Similarly we collect terms with base 2, base 3 and so on in the above equation. We have $$\begin{aligned} & \Rightarrow \sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}=\left( ^{m+1}{{C}_{1}}{{1}^{1}}{{+}^{m+1}}{{C}_{2}}{{1}^{2}}{{...}^{m+1}}{{C}_{m}}{{1}^{m}} \right)+\left( ^{m+1}{{C}_{1}}{{2}^{1}}{{+}^{m+1}}{{C}_{2}}{{2}^{2}}{{...}^{m+1}}{{C}_{m}}{{2}^{m}} \right)... \\\ & +\left( ^{m+1}{{C}_{1}}{{n}^{1}}{{+}^{m+1}}{{C}_{2}}{{n}^{2}}{{...}^{m+1}}{{C}_{m}}{{n}^{m}} \right).....(2) \\\ \end{aligned}$$ Let us observe the first set of summation$^{m+1}{{C}_{1}}{{1}^{1}}{{+}^{m+1}}{{C}_{2}}{{1}^{2}}{{...}^{m+1}}{{C}_{m}}{{1}^{m}}$. We see that it is almost a binomial expansion in the form ${{\left( 1+x \right)}^{a}}$, but it is missing the first term $^{m+1}{{C}_{0}}$ and the last term $^{m+1}{{C}_{m+1}}{{1}^{m+1}}$. So we use the equation (1) and substitute $x=1$ and $a=m+1$ to have $$\begin{aligned} & {{\left( 1+1 \right)}^{m+1}}-1-{{1}^{m+1}}{{=}^{m+1}}{{C}_{1}}{{1}^{1}}{{+}^{m+1}}{{C}_{1}}{{1}^{2}}...{{+}^{m+1}}{{C}_{m}}{{1}^{m}} \\\ & \Rightarrow {{\left( 2 \right)}^{m+1}}-1-{{1}^{m+1}}{{=}^{m+1}}{{C}_{1}}{{1}^{1}}{{+}^{m+1}}{{C}_{1}}{{1}^{2}}...{{+}^{m+1}}{{C}_{m}}{{1}^{m}} \\\ \end{aligned}$$ Similarly we observe the first second set of summation $^{m+1}{{C}_{1}}{{2}^{1}}{{+}^{m+1}}{{C}_{2}}{{2}^{2}}{{...}^{m+1}}{{C}_{m}}{{2}^{m}}$. We see that it is also almost a binomial expansion in the form ${{\left( 1+x \right)}^{a}}$, but it is missing the first term $^{m+1}{{C}_{0}}$ and the last term $^{m+1}{{C}_{m+1}}{{2}^{m+1}}$. So we use the equation (1) and substitute $x=2$ and $a=m+1$ to have $$\begin{aligned} & {{\left( 1+2 \right)}^{m+1}}-1-{{2}^{m+1}}{{=}^{m+1}}{{C}_{1}}{{2}^{1}}{{+}^{m+1}}{{C}_{1}}{{2}^{2}}...{{+}^{m+1}}{{C}_{m}}{{2}^{m}} \\\ & \Rightarrow {{\left( 3 \right)}^{m+1}}-1-{{2}^{m+1}}{{=}^{m+1}}{{C}_{1}}{{2}^{1}}{{+}^{m+1}}{{C}_{1}}{{2}^{2}}...{{+}^{m+1}}{{C}_{m}}{{2}^{m}} \\\ \end{aligned}$$ We can similarly substitute $x=3,4...,n$ and find expression of the form ${{\left( 1+x \right)}^{m+1}}-1-{{x}^{m+1}}$ . We replace the summations in equation (2) and get $$\begin{aligned} & \Rightarrow \sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}={{\left( 2 \right)}^{m+1}}-1-{{1}^{m+1}}+{{\left( 3 \right)}^{m+1}}-1-{{2}^{m+1}}+...+{{\left( n+1 \right)}^{m+1}}-1-{{n}^{m+1}} \\\ & \Rightarrow \sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}={{2}^{m+1}}-{{1}^{m+1}}+{{3}^{m+1}}-{{2}^{m+1}}+{{4}^{m+1}}-{{3}^{m+1}} \\\ & ...+{{\left( n+1 \right)}^{m+1}}-{{n}^{m+1}}-\left( 1+1+1...n\text{ times} \right) \\\ \end{aligned}$$ We observe that all terms except ${{1}^{m+1}}=1$ and ${{\left( n+1 \right)}^{m+1}}$ cancel out. So we have $$\begin{aligned} & \Rightarrow \sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}=-1+{{\left( n+1 \right)}^{m+1}}-n \\\ & \Rightarrow \sum\limits_{r=1}^{m}{^{m+1}{{C}_{r}}{{s}_{r}}}={{\left( n+1 \right)}^{m+1}}-\left( n+1 \right)...(3) \\\ \end{aligned}$$ Hence the statement is proved. We know from the formula of sum of first $n$, squared $n$ and cubed $n$ terms that $$\begin{aligned} & 1+2+3...n=\dfrac{n\left( n+1 \right)}{2}={{s}_{1}} \\\ & {{1}^{2}}+{{2}^{2}}+{{3}^{2}}...{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}={{s}_{2}} \\\ & {{1}^{3}}+{{2}^{3}}+{{3}^{3}}...{{n}^{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}={{s}_{3}} \\\ \end{aligned}$$ We now use the above values and putting $n=4$ in equation(3) and get, $$\begin{aligned} & \sum\limits_{r=1}^{4}{^{4+1}{{C}_{r}}{{s}_{r}}}={{\left( n+1 \right)}^{4+1}}-\left( n+1 \right) \\\ &{{\Rightarrow }^{5}}{{C}_{1}}{{s}_{1}}{{+}^{5}}{{C}_{2}}{{s}_{2}}{{+}^{5}}{{C}_{3}}{{s}_{3}}{{+}^{5}}{{C}_{4}}{{s}_{4}}={{\left( n+1 \right)}^{5}}-\left( n+1 \right) \\\ & \Rightarrow 5\dfrac{n\left( n+1 \right)}{2}+10\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+10\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+5{{s}_{4}}={{\left( n+1 \right)}^{5}}-\left( n+1 \right) \\\ & \Rightarrow 5{{s}_{4}}={{\left( n+1 \right)}^{5}}-\left( n+1 \right)-\dfrac{5n\left( n+1 \right)}{2}+\dfrac{5n\left( n+1 \right)\left( 2n+1 \right)}{3}+\dfrac{5{{n}^{2}}{{\left( n+1 \right)}^{2}}}{2} \\\ \end{aligned}$$ We take $\dfrac{n+1}{6}$ common form the right hand side and get, $$\begin{aligned} & \Rightarrow 5{{s}_{4}}={{\left( n+1 \right)}^{5}}-\left( n+1 \right)-\dfrac{5n\left( n+1 \right)}{2}-\dfrac{5n\left( n+1 \right)\left( 2n+1 \right)}{3}-\dfrac{5{{n}^{2}}{{\left( n+1 \right)}^{2}}}{2} \\\ & \Rightarrow 5{{s}_{4}}=\dfrac{n+1}{6}\left( 6{{\left( n+1 \right)}^{4}}-6-15n\left( n+1 \right)-15n\left( 2n+1 \right)-15{{n}^{2}}{{\left( n+1 \right)}^{2}} \right) \\\ & \Rightarrow 5{{s}_{4}}=\dfrac{n+1}{6}\left( 6{{n}^{4}}+9{{n}^{3}}+{{n}^{2}}-n \right) \\\ & \Rightarrow {{s}_{4}}=\dfrac{n\left( n+1 \right)}{30}\left( 6{{n}^{3}}+9{{n}^{2}}+n-1 \right) \\\ \end{aligned}$$ We further factorize $6{{n}^{3}}+9{{n}^{2}}+n-1$ to have the required value from the question as , $${{s}_{4}}=\dfrac{1}{30}n\left( n+1 \right)\left( 2n+1 \right)\left( 3{{n}^{2}}+6n-1 \right)$$ **Note:** We note that the binomial expansion of ${{\left( 1+x \right)}^{n}}$ is specialized form of binomial expansion of ${{\left( x+y \right)}^{n}}$ which is given as $\left( ^{n}{{C}_{0}} \right){{x}^{n}}{{y}^{0}}+\left( ^{n}{{C}_{1}} \right){{x}^{n-1}}y+...+\left( ^{n}{{C}_{n}} \right){{x}^{0}}{{y}^{n}}$ where $n$ is always a non-negative integers and $x,y$ are real numbers. We also note that the obtained expression of ${{s}_{4}}$ is the sum of first $n$ terms raised to the power of 4.