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  • For the expansion of , which of the following statement is true.<

For the expansion of \left ( x-\frac{1}{x} \right )^n, which of the following statement is true.

Option: 1

Middle term of the expansion is independent of x if n is even


Option: 2

Middle term of the expansion is dependent on x if n is even


Option: 3

There are two middle terms if n is even


Option: 4

None of these


Answers (1)

best_answer

We have learnt

The middle term in the expansion (x + y)n, depends on the value of 'n'. 

Case 1 When 'n' is even

Only one middle term : \left(\frac{\mathrm{n}}{2}+1\right)^{\mathrm{th}}term.

It is given by \mathrm{T} _{\frac{\mathrm{n}}{2}+1}=\left(\begin{array}{c}{\mathrm{n}} \\ {\frac{\mathrm{n}}{2}}\end{array}\right) \mathrm{x}^{\frac{\mathrm{n}}{2}} \mathrm{y}^{\frac{\mathrm{n}}{2}} .

Case 2 When 'n' is odd

Two middle terms in the expansion: \left(\frac{n+1}{2}\right)^{t h} \text { and }\left(\frac{n+3}{2}\right)^{t h} term.

Given by

\mathrm{T}_{\frac{\mathrm{n}+1}{2}}=\left(\begin{array}{c}{\mathrm{n}} \\ {\frac{\mathrm{n}-1}{2}}\end{array}\right) \mathrm{x}^{\frac{\mathrm{n}+1}{2}} \cdot \mathrm{y}^{\frac{\mathrm{n}-1}{2}} \quad ,\quad \mathrm{T}_{\frac{\mathrm{n}+3}{2}}=\left(\begin{array}{c}{\mathrm{n}} \\ {\frac{\mathrm{n}+1}{2}}\end{array}\right) \mathrm{x}^{\frac{\mathrm{n}-1}{2}} \cdot \mathrm{y}^{\frac{\mathrm{n}+1}{2}}

 

Now,

\large \left(x-\frac{1}{x}\right)^{n}=\sum_{r=0}^{n} \;^{n} C_{r} x^{n-r} \cdot\left(\frac{-1}{x}\right)^{r}

 If n is even middle term is 

\large T_{\frac{n}{2}+1}= \;^{n} C_{\frac{n}{2}}\; x^{n-\frac{n}{2}} \cdot\left(-\frac{1}{x}\right)^{\frac{n}{2}}\\T_{\frac{n}{2}+1}=^{n} C_{\frac{n}{2}}\; x^{n-\frac{n}{2}}\cdot x^{-\frac{n}{2}}(-1)^\frac{n}{2}

\large T_{\frac{n}{2}+1}= (-1)^\frac{n}{2}\;^{n} C_{\frac{n}{2}}\; x^{n-\frac{n}{2}-\frac{n}{2}}=(-1)^\frac{n}{2}\;^{n} C_{\frac{n}{2}}

This is independent of x, hence, option A is correct

Posted by

HARSH KANKARIA

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