Question

If $r$th and $\left(r+1\right)$th terms in the expansion of ${\left(p+q\right)}^n$ are equal, then the value of $\frac{(n+1)q}{r(p+q)}$ is

• A. $0$
• B. $1$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B

$1$

Apply $r$th term that is ${\mathit{T}}_{\mathbf{r}}\mathbf{=}{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}\mathbf{-}\mathbf{1}}{\mathit{p}}^{\mathbf{n}\mathbf{-}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathit{q}}^{\mathbf{r}\mathbf{-}\mathbf{1}}$for the further expansion:

For the expansion of ${\left(p+q\right)}^n$ the general term is given by $T_{r+1}={}^{n}C_r{p}^{n-r}{q}^r$

We know ${}^{n}C_r=\frac{n!}{(n-r)!r!}$,

Given that $r$th and $\left(r+1\right)$th terms are equal then equate the terms as follows:

$\begin{array}{l}\therefore T_r=T_{r+1}\\ \Rightarrow C_{r-1}{p}^{n-(r-1)}{q}^{r-1}={}^{n}C_r{p}^{n-r}{q}^r\\ \Rightarrow {}^{n}C_{r-1}{p}^{n-r+1}{q}^{r-1}={}^{n}C_r{p}^{n-r}{q}^r\\ \Rightarrow \frac{n!}{(n-\left(r-1\right))!(r-1)!}{p}^{n-r+1}{q}^{r-1}=\frac{n!}{(n-r)!\left(r\right)!}{p}^{n-r}{q}^r\\ \Rightarrow \frac{n!}{(n-r+1)!(r-1)!}\times \frac{(n-r)!\left(r\right)!}{n!}=\frac{p^{n-r}{q}^r}{p^{n-r+1}{q}^{r-1}}\\ \Rightarrow \frac{(n-r)!\left(r\right)(r-1)!}{(n-r+1)(n-r)!(r-1)!}=p^{n-r-(n-r+1)}{q}^{r-(r-1)}\\ \mathbf{\Rightarrow }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{r}{(n-r+1)}=p^{-1}q\\ \Rightarrow \frac{r}{(n-r+1)}=\frac{q}{p}\\ \Rightarrow pr=qn-qr+q\\ \Rightarrow pr+qr=qn+q\\ \Rightarrow (p+q)r=(n+1)q\\ \Rightarrow \frac{(n+1)q}{r(p+q)}=1\end{array}$

Hence, the correct option is (B).