Find the middle term in the expansion of $$(x-\dfrac{ 1}{x})^{2n+1}$$

Asked by Sakshi | 1 year ago |  338

##### Solution :-

where, n = (2n + 1) is an (odd number)

So the middle terms are $$\dfrac{(n+1)}{2}$$

= $$\dfrac{ (2n+1+1)}{2}$$

=$$\dfrac{ (2n+2)}{2}$$ = (n + 1) and

$$\dfrac{(n+1)}{2} + 1$$

=$$\dfrac{ (2n+1+1)}{2 + 1}$$

= (n + 1 + 1) = (n + 2)

The terms are (n + 1)th and (n + 2)th.

Now,

Tn = Tn+1 And,

Tn+2 = Tn+1+1 Hence, the middle term is (-1)n.2n+1Cn x and (-1)n+1.2n+1Cn ($$\dfrac{1}{x}$$).

Answered by Sakshi | 1 year ago

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