The positive value of for which the co-efficient of x^2 in the expression x^2( x+x^2)^10 i…

Question

The positive value of $\lambda$ for which the co-efficient of $x^2$ in the expression $x^2\left(\sqrt x+\dfrac{\lambda}{x^2}\right)^{10}$ is 720, is:

Accepted Answer

The general term in the expansion of the binomial expression $(a+b)^n$ is

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

Therefore, the general term in the expansion of the binomial expression

$x^2\left(\sqrt{x}+\frac{\lambda}{x^2}\right)^{10}$ is

$$
\begin{aligned}
T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\
& ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\
& ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r}
\end{aligned}
$$

Now, for the coefficient of $x^2$,

$$
\begin{aligned}
 2+\frac{10-r}{2}-2 r =2 \\
\Rightarrow  \frac{10-r}{2}-2 r =0 \\
\Rightarrow  10-r =4 r \Rightarrow r=2
\end{aligned}
$$

So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$

$$
\begin{aligned}
\Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\
\Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\
\Rightarrow & 45 \lambda^2 =720 \\
\Rightarrow & \lambda^2 =16 \\
\Rightarrow & \lambda = \pm 4
\end{aligned}
$$

Given the problem asks for the positive value of $\lambda$, $\lambda = 4$.

So, the correct option is (A) 4.

Solutions · 1