First slide

Binomial theorem for positive integral Index

Question

The positive value of $λ$ for which the coefficient of x² in the expression $x^{2} {(\sqrt{x} + \frac{λ}{x^{2}})}^{10}$ is 720 is

Difficult

A

3
B

$\sqrt{5}$
C

$2 \sqrt{2}$
D

4

Solution

The general term in the expansion of binomial expression $(a + b)^{n} is T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$
so the general term in the expansion of binomial expression $x^{2} {(\sqrt{x} + \frac{λ}{x^{2}})}^{10}$ is
$\begin{matrix} T_{r + 1} = x^{2} (^{10} C_{r} (\sqrt{x})^{10 - r} {(\frac{λ}{x^{2}})}^{r}) \\ =^{10} C_{r} x^{2} \cdot x^{\frac{10 - r}{2}} λ^{r} x^{- 2 r} \\ =^{10} C_{r} λ^{r} x^{2 + \frac{10 - r}{2} - 2 r} \end{matrix}$
Now, for the coefficient of x²
$\begin{array}{l} put, 2 + \frac{10 - r}{2} - 2 r = 2 \\ \Rightarrow \frac{10 - r}{2} - 2 r = 0 \\ \Rightarrow 10 - r = 4 r \Rightarrow r = 2 \end{array}$
So, the coefficient of $x^{2} {is}^{10} C_{2} λ^{2} = 720$
$\begin{array}{lrl} \Rightarrow \frac{10!}{2! 8!} λ^{2} = 720 \\ \Rightarrow \frac{10 \cdot 9 \cdot 8!}{2 \cdot 8!} λ^{2} = 720 \\ \Rightarrow 45 λ^{2} = 720 \\ \Rightarrow λ^{2} = 16 \Rightarrow λ = \pm 4 \end{array}$

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