Question

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

• A. 0
• B. 1
• C. 1/4
• D. 1/2

Answer 1

Answer: (B)

Given, $(p+q{)}^n$
$T_r=T_{(r-1)+1}={}^n{C}_{r-1}{p}^{n-r+1}\cdot q^{r-1}$
and $T_{r+1}={}^n{C}_r{p}^{n-r}\cdot q^r$
From question,
${}^n{C}_{r-1}{p}^{n-r+1}\cdot q^{r-1}={}^n{C}_r{p}^{n-r}\cdot q^r$
$\frac{n!}{(r-1)!(n-r+1)(n-r)!}\cdot p^{n-r}{q}^r\cdot \frac{p}{q}$
$=\frac{n!}{r(r-1)!(n-r)!}\cdot p^{n-r}\cdot q^r$
$\Rightarrow \frac{1}{(n-r+1)}\cdot \frac{p}{q}=\frac{1}{r}$
$\Rightarrow pr=qn-qr+q$
$\Rightarrow \frac{q(n+1)}{r(p+q)}=1$