Question

If the $4^{th}$ terms in the expansion of ${(px+\frac{1}{x})}^n$ is dependent of $x$, find the value of $n$. Also calculate $p$ if the $4^{th}$ term is $5/2$.

Answer 1

$(px+\frac{1}{x}{)}^n$

a = px $b=\frac{1}{x}=x^{-1}$

General term

$T_{r+1}{=}^n{C}_r{a}^r{b}^{n-r}$

${=}^n{C}_r\left(px{)}^r\right(x^{-1}{)}^{n-r}$

$4^{th}$ term is independent of x

i e when r = 3 power of x = 0

$\therefore T_{3+1}{=}^n{C}_3{p}^3{x}^3(x^{-1}{)}^{n-3}$

${=}^n{C}_3{p}^3{x}^3{x}^{3-n}$

${=}^n{C}_3{p}_3{x}_{6-n}$

$\Rightarrow 6-n=0$ $\therefore n=6$

Also, $4^{th}$ term is $\frac{5}{2}$ (given)

${\therefore }^6{C}_3{p}^3=\frac{5}{2}$

$\frac{6\times 5\times 4}{3\times 2\times 1}{p}^3=\frac{5}{2}$

$\therefore p^3=\frac{1}{8}$ $\therefore p=\frac{1}{2}$