Question

If in the expansion of $(1+px{)}^n,n\in N$, the coefficient of $x$ and $x^2$ are $8$ and $24$, then

• A. $n=3,p=2$
• B. $n=5,p=3$
• C. $n=4,p=3$
• D. $n=4,p=2$

Answer 1

Answer: (D)

We have, $(1+px{)}^n$
$\therefore T_{r+1}={}^n{C}_r(1{)}^{n-r}(px{)}^r$
$\Rightarrow T_{r+1}={}^n{C}_r{p}^r{x}^r$
Coefficient of $x^r={}^n{C}_r{p}^r$
Now coefficient of $x={}^n{C}_{1}p=8$ [put $r$ = 1]
$\Rightarrow np=8$ ....(i)
Also, coefficient of $x^2={}^n{C}_2{p}^2=24$ [put $r$ = 2]
$\Rightarrow \frac{n(n-1)}{2}{p}^2=24$
$\Rightarrow \frac{n(n-1)}{2}{\left(\frac{8}{n}\right)}^2=24$ (from (i))
$\Rightarrow \frac{64(n-1)}{2n}=24\Rightarrow 4(n-1)=3n\Rightarrow n=4$
From (i), we get $p=2$