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,
${ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}$
$\frac{n !}{(r-1) !(n-r+1)(n-r) !} \cdot p^{n-r} q^{r} \cdot \frac{p}{q}$
$=\frac{n !}{r(r-1) !(n-r) !} \cdot p^{n-r} \cdot q^{r}$
$\Rightarrow \frac{1}{(n-r+1)} \cdot \frac{p}{q}=\frac{1}{r}$
$\Rightarrow p r=q n-q r+q$
$\Rightarrow \frac{q(n+1)}{r(p+q)}=1$