Question

The sum $\sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}$ is equal to
(a) $\dfrac{\left( n+2 \right)\left( 2n-1 \right)!}{n!\left( n-1 \right)!}$
(b) $\dfrac{\left( n+2 \right)\left( 2n+1 \right)!}{n!\left( n-1 \right)!}$
(c) $\dfrac{\left( n+2 \right)\left( 2n+1 \right)!}{n!\left( n+1 \right)!}$
(d) $\dfrac{\left( n+2 \right)\left( 2n-1 \right)!}{n!\left( n+1 \right)!}$

Answer 1

We will first expand the term $\left( r+1 \right){}^{n}C_{r}^{2}$ by using distribution law of multiplication. After that we calculate each term in the expression by substituting the value ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Now we will get the simple values of each term of $\left( r+1 \right){}^{n}C_{r}^{2}$. So according to the problem we will apply the summation to get the result.

Complete step-by-step answer:
Given that, $\sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}$
Using the distribution law of multiplication to simplify the term $\left( r+1 \right){}^{n}C_{r}^{2}$, then we will get
$\left( r+1 \right){}^{n}C_{r}^{2}=r.{}^{n}C_{r}^{2}+{}^{n}C_{r}^{2}$
So, we can write the term $\sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}$ as $\sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\sum\limits_{r=0}^{n}{r.}{}^{n}C_{r}^{2}+\sum\limits_{r=0}^{n}{{}^{n}C_{r}^{2}}$
Now we will calculate the values of $r.{}^{n}C_{r}^{2}$ and ${}^{n}C_{r}^{2}$ individually, after that we will substitute them in the above equation.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, then the value of $r.{}^{n}{{C}_{r}}$ is calculated as
$r.{}^{n}{{C}_{r}}=r.\dfrac{n!}{r!\left( n-r \right)!}$
Now we are going to write $n!=n.\left( n-1 \right)!$ and $r!=r.\left( r-1 \right)!$ in the above equation, then
$r.{}^{n}{{C}_{r}}=r.\dfrac{n.\left( n-1 \right)!}{r.\left( r-1 \right)!\left( n-r \right)!}$
Now add and subtract one in the term $\left( n-r \right)!$, then we will have
$\begin{aligned} & r.{}^{n}{{C}_{r}}=n.\dfrac{\left( n-1 \right)!}{\left( r-1 \right)!\left[ n-r+1-1 \right]!} \\\ & \Rightarrow r.{}^{n}{{C}_{r}}=n.\dfrac{\left( n-1 \right)!}{\left( r-1 \right)!\left[ \left( n-1 \right)-\left( r-1 \right) \right]!} \\\ \end{aligned}$
We can write the term $\dfrac{\left( n-1 \right)!}{\left( r-1 \right)!\left[ \left( n-1 \right)-\left( r-1 \right) \right]!}$as ${}^{n-1}{{C}_{r-1}}$, then we will have
$r.{}^{n}{{C}_{r}}=n.{}^{n-1}{{C}_{r-1}}$
So from the value of $r.{}^{n}{{C}_{r}}$, we can write the$r.{}^{n}C_{r}^{2}=n.{}^{n-1}C_{r-1}^{2}$
$\begin{aligned} & \therefore \sum\limits_{r=0}^{n}{r.{}^{n}C_{r}^{2}}=\sum\limits_{r=0}^{n}{n.{}^{n-1}C_{r-1}^{2}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{r.{}^{n}C_{r}^{2}}=n\left[ {}^{n-1}C_{0}^{2}+{}^{n-1}C_{1}^{2}+{}^{n-1}C_{2}^{2}+...+{}^{n-1}C_{n-1}^{2} \right] \\\ \end{aligned}$
We have an identity which states that ${}^{m+n}{{C}_{k}}=\sum\limits_{r=0}^{k}{{}^{m}{{C}_{r}}\cdot {}^{n}{{C}_{k-r}}}$. Now, in this identity, we will substitute $m=k=n$. So, we get ${}^{n+n}{{C}_{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}\cdot {}^{n}{{C}_{n-r}}}$. We already know that ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$. Therefore, we get the following,
${}^{2n}{{C}_{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}^{2}}$
Hence, we have $\sum\limits_{r=0}^{n}{{}^{n}C_{r}^{2}}=\dfrac{2n!}{n!n!}.....\left( \text{i} \right)$ from this the value of $\sum\limits_{r=0}^{n}{{}^{n-1}C_{r}^{2}}=\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}$
$\Rightarrow \sum\limits_{r=0}^{n}{r.{}^{n}C_{r}^{2}}=n.\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}........\left( \text{ii} \right)$
Now adding the equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$, then we will get
$\begin{aligned} & \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=n.\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}+\dfrac{2n!}{n!n!} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=n.\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}+\dfrac{2n\left( 2n-1 \right)!}{n\left( n-1 \right)!n!} \\\ \end{aligned}$
Now taking $\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}$ as common in the above equation, then we will have
$\begin{aligned} & \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}\left[ n+\dfrac{2n}{n} \right] \\\ & \Rightarrow \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\dfrac{\left( n+2 \right)\left( 2n-1 \right)!}{\left( n-1 \right)!n!} \\\ \end{aligned}$

So, the correct answer is “Option A”.

Note: We can also solve the problem by using the formula $aC_{0}^{2}+\left( a+d \right)C_{1}^{2}+\left( a+2d \right)C_{2}^{2}+...+\left( a+nd \right)C_{n}^{2}=\left( \dfrac{2a+nd}{2} \right){}^{2n}{{C}_{n}}$
Substitute the value of $a=1$ and $d=1$, then
$\begin{aligned} & \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\left( \dfrac{2\times 1+n\times 1}{2} \right){}^{2n}{{C}_{n}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\left( \dfrac{n+2}{2} \right)\times \dfrac{2n!}{n!\times n!} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\dfrac{n+2}{2}\times 2n\times \dfrac{\left( 2n-1 \right)!}{n\times \left( n-1 \right)!\times n!} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{\left( r+1 \right){}^{n}C_{r}^{2}}=\dfrac{\left( n+2 \right)\times \left( 2n-1 \right)!}{n!\times \left( n-1 \right)!} \\\ \end{aligned}$
From both the methods we get the same answer.