Question

Show that the middle term in the expansion of `(x - 1/x)^(2x)` is `(1 xx 3 xx 5 xx ... (2n - 1))/(n!) xx (-2)^n`

Answer 1

Given expression is `(x - 1/x)^(2x)`

Number of terms = 2n + 1 (odd)

∴ Middle term = `(2"n" + 1 + 1)/2` th term

i.e., (n + 1)^th term

General Term `"T"_(r + 1) = ""^n"C"_r (x)^(n - r)  (y)^r`

∴ `"T"_(n + 1) = ""^(2n)"C"_n  (x)^(2n - n) (-1/x)^n`

= `""^(2n)"C"_n (x)^n (-1)^n * 1/x^n`

= `(1)^n * ""^(2n)"C"_n`

= `(-1)^n * (2n!)/(n!(2n - n)!)`

= `(-1)^n * (2n!)/(n1n!)`

= `(-1)^n * (2n(2n - 1)(2n - 2)(2n - 3) ... 1)/(n!n(n - 1)(n - 2)(n - 3) ....1)`

= `(-1)^n (2n*(2n - 1)*2(n - 1)(2n - 3) ... 1)/(n!n(n - 1)(n - 2)(n - 3) ....1)`

= `((-1)^n * 2^n * [n(n - 1)(n - 2) ...] * [(2n - 1) * (2n - 3) ... 5 * 3*1])/(n! * n(n - 1)(n - 2)(n - 3) ...1)`

= `((-2)^n[(2n - 1)(2n - 3) ... 5*3*1])/(n!)`

= `(1 xx 3 xx 5 xx ... (2n - 1))/(n!) xx (-2)^n`

Hence, the middle term = `(1 xx 3 xx 5 xx ... (2n - 1))/(n!) xx (-2)^n`