Question

Suppose det $\left[ \begin{matrix} \sum\limits_{k=0}^{n}{k} & \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}} \\\ \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k} & \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{3}^{k}}} \\\ \end{matrix} \right]=0$ holds for some positive integer $n$. Then $\sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}$ is equal to ?

Answer 1

Usually when we have a determinant or det , we first have to cross multiply or simplify it according to the rows and columns. Since it is $2\times 2$ determinant , let us just cross-multiply. The question says that $\left[ \begin{matrix} \sum\limits_{k=0}^{n}{k} & \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}} \\\ \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k} & \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{3}^{k}}} \\\ \end{matrix} \right]=0$ holds good for some positive integer $n$ . Let us first find that particular value of $n$ for which it holds good. And then we can just expand the summation and substitute the value of $n$ to find the value of $\sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}$.

Complete step-by-step solution:
We know that the sum of first $n$ natural numbers is $\dfrac{n\left( n+1 \right)}{2}$
$\Rightarrow 1+2+3+4.........n=\dfrac{n\left( n+1 \right)}{2}$….$eqn\left( 1 \right)$
Now let us expand the first summation i.e $\sum\limits_{k=0}^{n}{k}$.
Upon expanding we get the following :

$\Rightarrow \sum\limits_{k=0}^{n}{k}=0+1+2+3+.......n$
From $eqn\left( 1 \right)$, we can infer :
$\begin{aligned} & \Rightarrow \sum\limits_{k=0}^{n}{k}=0+1+2+3+.......n. \\\ & \Rightarrow \sum\limits_{k=0}^{n}{k}=\dfrac{n\left( n+1 \right)}{2}.....result(1) \\\ \end{aligned}$
Let us expand ${{\left( 1+x \right)}^{n}}$ using binomial theorem.
$\Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+........{}^{n}{{C}_{n}}{{x}^{n}}$ .
Now let us differentiate on both the sides.
$\begin{aligned} & \Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+........{}^{n}{{C}_{n}}{{x}^{n}} \\\ & \Rightarrow n{{\left( 1+x \right)}^{n-1}}=0+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}\left( 2x \right)+{}^{n}{{C}_{3}}\left( 3{{x}^{2}} \right)+.......+{}^{n}{{C}_{n}}{{x}^{n-1}} \\\ \end{aligned}$
Now let us substitute $x=1.$
$\begin{aligned} & \Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+........{}^{n}{{C}_{n}}{{x}^{n}} \\\ & \Rightarrow n{{\left( 1+x \right)}^{n-1}}=0+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}\left( 2x \right)+{}^{n}{{C}_{3}}\left( 3{{x}^{2}} \right)+.......+{}^{n}{{C}_{n}}\left( n{{x}^{n-1}} \right) \\\ & \Rightarrow n{{\left( 1+1 \right)}^{n-1}}={}^{n}{{C}_{1}}+2{}^{n}{{C}_{2}}+3{}^{n}{{C}_{3}}+4{}^{n}{{C}_{4}}+........+n{}^{n}{{C}_{n}} \\\ & \Rightarrow {{2}^{n-1}}n={}^{n}{{C}_{1}}+2{}^{n}{{C}_{2}}+3{}^{n}{{C}_{3}}+4{}^{n}{{C}_{4}}+........+n{}^{n}{{C}_{n}}.....eqn(2) \\\ \end{aligned}$
Now let us expand and solve the second summation i.e $\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k}$.
$\Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k}={}^{n}{{C}_{0}}0+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}2+{}^{n}{{C}_{3}}3+........{}^{n}{{C}_{n}}n.$
From $eqn(2)$ we can infer that :
$\begin{aligned} & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k}={}^{n}{{C}_{0}}0+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}2+{}^{n}{{C}_{3}}3+........{}^{n}{{C}_{n}}n. \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k}={{2}^{n-1}}n......result(2) \\\ \end{aligned}$
Now let us solve the next summation i.e $\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}$.
We can write $r{}^{n}{{C}_{r}}$ as $n{}^{n-1}{{C}_{r-1}}$ as both are same.
$\begin{aligned} & \Rightarrow r{}^{n}{{C}_{r}}=r\dfrac{n!}{r!\left( n-r \right)!} \\\ & \Rightarrow r{}^{n}{{C}_{r}}=r\dfrac{n(n-1)!}{r(r-1)!\left( n-r \right)!} \\\ & \Rightarrow r{}^{n}{{C}_{r}}=\dfrac{n(n-1)!}{(r-1)!\left( n-r \right)!} \\\ & \Rightarrow r{}^{n}{{C}_{r}}=n{}^{n-1}{{C}_{r-1}} \\\ & \Rightarrow {{r}^{2}}{}^{n}{{C}_{r}}=r.r.{}^{n}{{C}_{r}} \\\ & \Rightarrow {{r}^{2}}{}^{n}{{C}_{r}}=rn{}^{n-1}{{C}_{r-1}} \\\ & \Rightarrow {{r}^{2}}{}^{n}{{C}_{r}}=n\left( \left( r-1 \right)+1 \right){}^{n-1}{{C}_{r-1}} \\\ & \Rightarrow {{r}^{2}}{}^{n}{{C}_{r}}=n\left( r-1 \right){}^{n-1}{{C}_{r-1}}+n{}^{n-1}{{C}_{r-1}} \\\ & \Rightarrow {{r}^{2}}{}^{n}{{C}_{r}}=n\left( n-1 \right){}^{n-2}{{C}_{r-2}}+n{}^{n-1}{{C}_{r-1}}.....eqn(3) \\\ \end{aligned}$
$r$ is just some random variable and it is not related to the question whatsoever.
We can infer from $eqn(3)$ that :

& \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=\sum\limits_{k=2}^{n}{n\left( n-1 \right){}^{n-2}{{C}_{k-2}}}+\sum\limits_{k=1}^{n}{n{}^{n-1}{{C}_{k-1}}} \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=n\left( n-1 \right)\sum\limits_{k=2}^{n}{{}^{n-2}{{C}_{k-2}}+}n\sum\limits_{k=1}^{n}{{}^{n-1}{{C}_{k-1}}} \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=n\left( n-1 \right)\left[ {}^{n-2}{{C}_{0}}+{}^{n-2}{{C}_{1}}+{}^{n-2}{{C}_{2}}+.......{}^{n-2}{{C}_{n-2}} \right]+n\left[ {}^{n-1}{{C}_{0}}+{}^{n-1}{{C}_{1}}+{}^{n-1}{{C}_{2}}+....+{}^{n-1}{{C}_{n-1}} \right] \\\ \end{aligned}$$ We know that $${}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+......{}^{n}{{C}_{n}}={{2}^{n}}$$ .So, $$\begin{aligned} & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=\sum\limits_{k=2}^{n}{n\left( n-1 \right){}^{n-2}{{C}_{k-2}}}+\sum\limits_{k=1}^{n}{n{}^{n-1}{{C}_{k-1}}} \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=n\left( n-1 \right)\sum\limits_{k=2}^{n}{{}^{n-2}{{C}_{k-2}}+}n\sum\limits_{k=1}^{n}{{}^{n-1}{{C}_{k-1}}} \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=n\left( n-1 \right)\left[ {}^{n-2}{{C}_{0}}+{}^{n-2}{{C}_{1}}+{}^{n-2}{{C}_{2}}+.......{}^{n-2}{{C}_{n-2}} \right]+n\left[ {}^{n-1}{{C}_{0}}+{}^{n-1}{{C}_{1}}+{}^{n-1}{{C}_{2}}+....+{}^{n-1}{{C}_{n-1}} \right] \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}}=n\left( n-1 \right){{2}^{n-2}}+n{{2}^{n-1}}......result(3) \\\ \end{aligned}$$ Now let us solve the last summation i.e $$\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{3}^{k}}}$$. Let us solve this. $\Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}}{{3}^{k}}=\left[ {}^{n}{{C}_{0}}+3{}^{n}{{C}_{1}}+{{3}^{2}}{}^{n}{{C}_{2}}+.......+{{3}^{n}}{}^{n}{{C}_{n}} \right]$ . This is same as substituting $x=3$ in the binomial expansion of ${{\left( 1+x \right)}^{n}}$ which results in $${{4}^{n}}$$ . $\Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+.......+{}^{n}{{C}_{n}}{{x}^{n}}$ Now substitute $x=3$. We get the following : $\begin{aligned} & \Rightarrow {{\left( 1+3 \right)}^{n}}=\left[ {}^{n}{{C}_{0}}+3{}^{n}{{C}_{1}}+{{3}^{2}}{}^{n}{{C}_{2}}+.......+{{3}^{n}}{}^{n}{{C}_{n}} \right]=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{3}^{k}}} \\\ & \Rightarrow \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{3}^{k}}}={{4}^{n}}......result(4) \\\ \end{aligned}$ Now let us substitute all these results which we got in the determinant. $$\begin{aligned} & \Rightarrow \left[ \begin{matrix} \sum\limits_{k=0}^{n}{k} & \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{k}^{2}}} \\\ \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}k} & \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{3}^{2}}} \\\ \end{matrix} \right]=0 \\\ & \Rightarrow \left[ \begin{matrix} \dfrac{n\left( n+1 \right)}{2} & n\left( n-1 \right){{2}^{n-2}}+n{{2}^{n-1}} \\\ {{2}^{n-1}}n & {{4}^{n}} \\\ \end{matrix} \right]=0 \\\ \end{aligned}$$ Now let us cross-multiply. $$\begin{aligned} & \Rightarrow \left[ \begin{matrix} \dfrac{n\left( n+1 \right)}{2} & n\left( n-1 \right){{2}^{n-2}}+n{{2}^{n-1}} \\\ {{2}^{n-1}}n & {{2}^{2n}} \\\ \end{matrix} \right]=0 \\\ & \Rightarrow \dfrac{n\left( n+1 \right){{2}^{2n}}}{2}-{{2}^{n-1}}n(n\left( n-1 \right){{2}^{n-2}}+n{{2}^{n-1}})=0 \\\ & \Rightarrow n\left( n+1 \right){{2}^{2n-1}}-{{2}^{n-1}}n(n\left( n-1 \right){{2}^{n-2}}+n{{2}^{n-1}})=0 \\\ \end{aligned}$$ Taking ${{2}^{n-1}}n$ common, $$\begin{aligned} & \Rightarrow n\left( n+1 \right){{2}^{2n-1}}-{{2}^{n-1}}n(n\left( n-1 \right){{2}^{n-2}}+n{{2}^{n-1}})=0 \\\ & \Rightarrow {{2}^{n-1}}n\left[ {{2}^{n}}\left( n+1 \right)-{{n}^{2}}{{2}^{n-2}}+n{{2}^{n-2}}-n{{2}^{n-1}} \right]=0 \\\ & \Rightarrow {{2}^{n-1}}n\left[ {{2}^{n}}n+{{2}^{n}}-{{n}^{2}}{{2}^{n-2}}+n{{2}^{n-2}} \right]=0 \\\ & \Rightarrow {{2}^{2n-1}}n\left[ n+1-\dfrac{{{n}^{2}}}{4}-\dfrac{n}{4} \right]=0 \\\ & \Rightarrow {{2}^{2n-1}}n\left[ \dfrac{-{{n}^{2}}+3n+4}{4} \right]=0 \\\ & \Rightarrow {{2}^{2n-1}}n\left[ -{{n}^{2}}+3n+4 \right]=0 \\\ \end{aligned}$$ Now let us split the middle term to find the value of $n$ . Upon doing so, we get the following : $$\begin{aligned} & \Rightarrow {{2}^{2n-1}}n\left[ -{{n}^{2}}+4n-n+4 \right]=0 \\\ & \Rightarrow {{2}^{2n-1}}n\left[ n\left( -n+4 \right)+1\left( -n+4 \right) \right]=0 \\\ & \Rightarrow {{2}^{2n-1}}n\left[ \left( n+1 \right)\left( -n+4 \right) \right]=0 \\\ \end{aligned}$$ So here $n$ has three values. $0$ or $-1$ or $4$ .$n$can’t be $0$ as $n$ has to be some non-zero value or else the answer would be $0$. $n$ can’t be a negative value. So the value of $n$is 4. Now let us solve $\sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}$. Let us look at the binomial expansion of ${{\left( 1+x \right)}^{n}}$. $\Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+........{}^{n}{{C}_{n}}{{x}^{n}}$ . Integrating on both sides. $\begin{aligned} & \Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+........{}^{n}{{C}_{n}}{{x}^{n}} \\\ & \Rightarrow \dfrac{{{\left( 1+x \right)}^{n+1}}}{n+1}={}^{n}{{C}_{0}}x+\dfrac{{}^{n}{{C}_{1}}{{x}^{2}}}{2}+\dfrac{{}^{n}{{C}_{2}}{{x}^{3}}}{3}+......+\dfrac{{}^{n}{{C}_{n}}{{x}^{n+1}}}{n+1} \\\ & \\\ \end{aligned}$ . Put $x=1$ . $\begin{aligned} & \Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+........{}^{n}{{C}_{n}}{{x}^{n}} \\\ & \Rightarrow \dfrac{{{\left( 1+x \right)}^{n+1}}}{n+1}={}^{n}{{C}_{0}}x+\dfrac{{}^{n}{{C}_{1}}{{x}^{2}}}{2}+\dfrac{{}^{n}{{C}_{2}}{{x}^{3}}}{3}+......+\dfrac{{}^{n}{{C}_{n}}{{x}^{n+1}}}{n+1} \\\ & \Rightarrow \dfrac{{{\left( 1+1 \right)}^{n+1}}}{n+1}={}^{n}{{C}_{0}}1+\dfrac{{}^{n}{{C}_{1}}{{1}^{2}}}{2}+\dfrac{{}^{n}{{C}_{2}}{{1}^{3}}}{3}+......+\dfrac{{}^{n}{{C}_{n}}{{1}^{n+1}}}{n+1} \\\ & \Rightarrow \dfrac{{{2}^{n+1}}}{n+1}={}^{n}{{C}_{0}}+\dfrac{{}^{n}{{C}_{1}}}{2}+\dfrac{{}^{n}{{C}_{2}}}{3}+......+\dfrac{{}^{n}{{C}_{n}}}{n+1}....eqn(4) \\\ \end{aligned}$ Let us expand $\sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}$. $\Rightarrow \sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}=\dfrac{{}^{n}{{C}_{0}}}{1}+\dfrac{{}^{n}{{C}_{1}}}{2}+\dfrac{{}^{n}{{C}_{2}}}{3}+.....+\dfrac{{}^{n}{{C}_{n}}}{n+1}$ From $eqn(4)$ we can infer : $\begin{aligned} & \Rightarrow \sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}=\dfrac{{}^{n}{{C}_{0}}}{1}+\dfrac{{}^{n}{{C}_{1}}}{2}+\dfrac{{}^{n}{{C}_{2}}}{3}+.....+\dfrac{{}^{n}{{C}_{n}}}{n+1} \\\ & \Rightarrow \sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}=\dfrac{{{2}^{n+1}}}{n+1} \\\ \end{aligned}$ From solving the determinant , the value of $n$ we got is $4$ . Let us substitute it here. $\Rightarrow \sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}=\dfrac{{{2}^{n+1}}}{n+1}=\dfrac{{{2}^{5}}}{5}=\dfrac{32}{5}=6.4$ . **$\therefore $ Hence the value of $\sum\limits_{k=0}^{n}{\dfrac{{}^{n}{{C}_{k}}}{k+1}}$is $6.4$ .** **Note:** This is a multi-concept question. We should be very careful with these kinds of questions as there is a lot of scope for making mistakes. We should be thorough with all the formulae such as sum of first $n$ natural numbers and all the other binomial theorem related formulae to complete the questions quickly. Box up all the results and equations so that there is no confusion. To simplify the question, we can just find out the value of $n$ by hit or trial method.