# Show that the middle term in the expansion of (x-1x)2x is 1×3×5×...(2n-1)n!×(-2)n - Mathematics

Sum

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

#### Solution

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

Concept: General and Middle Terms
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 13 | Page 143

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