Discussion Forum

In the expansion of $(1+x)^{n}$ , the coefficients of p th and (p+1) th terms are respectively p and q then p+q= 

Answer:

Explanation:

In the expansion of $(1+x)^{n}$ , the general term i.e, (r+1) th term is

  $T_{r+1}=^{n}C_{r}  (1)^{n-r} x^{r}=^{n}C_{r} x^{r}$

$\therefore$ Coefficient of (r+1) th term is $^{n}C_{r}$

Similarly , coefficent of pth term =$^{n}C_{p-1}$

$\therefore$           $p=^{n}C_{p-1}$(given)

            $p=\frac{n!}{(p-1)!(n-p+1)!}$   ........(i)

and coefficient of (p+1) th term= $^{n}C_{p}$

$\therefore$    $q=^{n}C_{p}$(given)     .........(ii)

On dividing Eq.(i) by Eq.(ii), we get

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

$=\frac{(p!)(n-p)!}{(n-p+1)!(p-1)!}=\frac{p(p-1)!(n-p)!}{(n-p+1)(n-p)!(p-1)!}$

$\Rightarrow$    $\frac{p}{q}=\frac{p}{n-p+1}$

$\Rightarrow$   $\frac{1}{q}=\frac{1}{n-p+1}$

$\Rightarrow$  $n-p+1=q$

$\Rightarrow$   $p+q=n+1$