Question

If ${(x^2+\frac{1}{x})}^n$ has exactly one middle term which is equal to $\alpha .x^3$ then the value of $(\alpha +n)$ is- ($n\in N$)

• A. $18$
• B. $21$
• C. $24$
• D. $26$

Answer 1

The correct option is D $26$

$\because$ Only one middle term $\Rightarrow$ $n$ is even middle term $=\alpha .x^3$
${{}^{n}C}_{n/2}{\left(x^2\right)}^{n/2}.{\left(\frac{1}{x}\right)}^{n/2}=\alpha .x^3$
${{}^{n}C}_{n/2}{x}^{n/2}=\alpha .x^3\Rightarrow \frac{n}{2}=3$
and $\alpha ={{}^{n}C}_{n/2}$
$\Rightarrow$ $n=6$ and $\alpha ={{}^{6}C}_3=20$

$\therefore \alpha +n=20+6=26$