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Mathematics Question on Binomial theorem

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

A

0

B

1

C

44565

D

44563

Answer

1

Explanation

Solution

Given, (p+q)n(p +q)^{n}
Tr=T(r−1)+1=nCr−1pn−r+1⋅qr−1T_{r}=T_{(r-1)+1}={ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}
and Tr+1=nCrpn−r⋅qrT_{r+1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}
From question,
nCr−1pn−r+1⋅qr−1=nCrpn−r⋅qr{ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}
n!(r−1)!(n−r+1)(n−r)!⋅pn−rqr⋅pq\frac{n !}{(r-1) !(n-r+1)(n-r) !} \cdot p^{n-r} q^{r} \cdot \frac{p}{q}
=n!r(r−1)!(n−r)!⋅pn−r⋅qr=\frac{n !}{r(r-1) !(n-r) !} \cdot p^{n-r} \cdot q^{r}
⇒1(n−r+1)⋅pq=1r\Rightarrow \frac{1}{(n-r+1)} \cdot \frac{p}{q}=\frac{1}{r}
⇒pr=qn−qr+q\Rightarrow p r=q n-q r+q
⇒q(n+1)r(p+q)=1\Rightarrow \frac{q(n+1)}{r(p+q)}=1