InterviewSolution
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InterviewSolution
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Show that the middle term in the expansion of `(1+x)^(2n)i s((1. 3. 5 (2n-1)))/(n !)2^n x^n ,w h e r en`is a positive integer.A. `(1* 3*5…(2n -1))/(n!) 2^(n) . X^(n)`B. `(1 * 3 * 5 …(2n -1))/(n!!) `C. `""^(2n)C(n)`D. `""^(n)C_(n-1) x^(n-1)` |
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Answer» Correct Answer - a We given expansion is `(1 +x)^(2n)` . Here , the index 2n is even So, `((2n)/(2) +1)^(th)` i.e. `(n+1)^(th)` term is the midle term `therefore ` Middle = `T_(n+1)` =` ""^(2n)C_(n) (1)^(2n-n) x^(n) = ""^(2n)C_(n) x^(n) = ((2n)!)/((2n -n)!n!) x^(n)` `=(1*2*3*4*5*6...(2n -3)(2n -2)(2n-2)(2n-1) (2n))/(n!n!)x^(n)` `=({1*3*5...(2n -3)(2n -1)}{*2*4*6...(2n-2)(2n)})/(n!n!)x^(n)` `=({1*3*5...(2n -3)(2n -1)}n!.2^(n))/(n!n!)x^(n)` `=(1*3*5...(2n -1))/(n!n!)2^(n)x^(n)`. |
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