First slide

Binomial theorem for positive integral Index

Question

Suppose $k \in R$ . Let $a$ be the coefficient of the middle term in the expansion of ${(\frac{k}{x} + \frac{x}{k})}^{10},$ and $b$ be the term independent of $x$ in the expansion of ${(\frac{k^{2}}{x} + \frac{x}{k})}^{10} .$ If $\frac{a}{b} = 1,$ then $k$ is equal to

Moderate

A

1
B

2
C

-3
D

any non-zero number

Solution

Middle term of ${(\frac{k}{x} + \frac{x}{k})}^{10}$ is $(\frac{10}{2} + 1)$ th = $6 t h$ term. We have
$\begin{array}{l} T_{6} =^{10} C_{5} {(\frac{k}{x})}^{10 - 5} {(\frac{x}{k})}^{3} =^{10} C_{5} \\ ∴ a =^{10} C_{5} \end{array}$
Also, $(r + 1) t h$ term of ${(\frac{k}{x} + \frac{x}{k})}^{10}$ is
$\begin{aligned} T_{r + 1} =^{10} C_{r} {(\frac{k^{2}}{x})}^{10 - r} {(\frac{x}{k})}^{r} \\ =^{10} C_{r} k^{20 - 3 r} x^{2 r - 10} \end{aligned}$
For term independent of $x,$ set $2 r - 10 = 0$ or $r = 5$ . Thus, term independent of $x$ in the expansion of ${(\frac{k^{2}}{x} + \frac{x}{k})}^{10}$ is $^{10} C_{5} k^{5} = b .$
As, $\frac{a}{b} = 1,$ we get $k^{5} = 1$ or $k = 1$

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