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Show that the middle term in the expansi...

Show that the middle term in the expansion `(x-1/x)^(2n)i s` `(1. 3. 5 (2n-1))/n(-2)^n` .

Text Solution

Verified by Experts

Given, expansion is `(x - (1)/(x))^(2n)`. This Binomial expansion has even power. So, this has one middle term.
i.e., `((2n)/(2) + 1)` th term `= (n + 1) th` term
`T_(n + 1) = .^(2n)C_(n) (x)^(2n - n) (-(1)/(x))^(n) = .^(2n)C_(n) x^(n) (-1)^(n) x^(-n)`
`= .^(2n)C_(n) (-1)^(n) = (-1)^(n) ((2n)!)/(n!n!) = (1.2.3.4.5 ..... (2n - 1) (2n))/(n!n!) (-1)^(n)`
`= (1.3.5 ... (2n - 1) .2.4.6.... (2n))/(12.3 .... n(n!)) (-1)^(n)`
`= (1.3.5....(2n - 1) . 2^(n) (1.2.3 .... n) (-1)^(n))/((1.2.3....n) (n!))`
`= ([1.3.5...(2n -1)])/(n!) (-2)^(n)`
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