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Question Number 18759 by 786786AM last updated on 29/Jul/17

The coefficient of the middle term in the  expansion of (1+x)^(2n)  is

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \:\mathrm{is} \\ $$

Answered by anuragbhandari123@gmail.com last updated on 29/Jul/17

((2nC(n+1)xofpower(n+1))/)

$$\frac{\mathrm{2}{nC}\left({n}+\mathrm{1}\right){xofpower}\left({n}+\mathrm{1}\right)}{} \\ $$

Answered by Tinkutara last updated on 30/Jul/17

Number of terms = 2n + 1  Middle term = (n + 1)^(th)  term  T_(n+1)  =^(2n) C_n x^n   Coefficient of middle term =^(2n) C_n   = (((2n)!)/(n! n!)) = ((1.2.3.4....(2n−2)(2n−1)(2n))/(n! n!))  = (([1.3.5....(2n−1)][2.4.6....(2n)])/(n! n!))  = (([1.3.5....(2n−1)]2^n [1.2.3....(n)])/(n! n!))  = ((1.3.5....(2n − 1))/(n!)) 2^n

$$\mathrm{Number}\:\mathrm{of}\:\mathrm{terms}\:=\:\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$$\mathrm{Middle}\:\mathrm{term}\:=\:\left({n}\:+\:\mathrm{1}\right)^{\mathrm{th}} \:\mathrm{term} \\ $$$${T}_{{n}+\mathrm{1}} \:=\:^{\mathrm{2}{n}} {C}_{{n}} {x}^{{n}} \\ $$$$\mathrm{Coefficient}\:\mathrm{of}\:\mathrm{middle}\:\mathrm{term}\:=\:^{\mathrm{2}{n}} {C}_{{n}} \\ $$$$=\:\frac{\left(\mathrm{2}{n}\right)!}{{n}!\:{n}!}\:=\:\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}....\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)}{{n}!\:{n}!} \\ $$$$=\:\frac{\left[\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}−\mathrm{1}\right)\right]\left[\mathrm{2}.\mathrm{4}.\mathrm{6}....\left(\mathrm{2}{n}\right)\right]}{{n}!\:{n}!} \\ $$$$=\:\frac{\left[\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}−\mathrm{1}\right)\right]\mathrm{2}^{{n}} \left[\mathrm{1}.\mathrm{2}.\mathrm{3}....\left({n}\right)\right]}{{n}!\:{n}!} \\ $$$$=\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)}{{n}!}\:\mathrm{2}^{{n}} \\ $$

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