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प्रश्न
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`
उत्तर
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`
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