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

Show that the middle term in the expansion of\left ( x-\frac{1}{x} \right )^{2n}is\frac{1\times 3 \times 5\times...........\times\left ( 2n-1 \right ) }{n!} \times\left ( -2 \right )^{n}

 

Answers (1)

Given: (x – 1/x)2n

Now, here the index = 2n(even)

There’s only one middle term, i.e., (2n/2 + 1)th term, viz., (n + 1)th term

Tn+1  = 2nCn(x)2n-n(-1/x)n

                 = 2nCn(-1)n

                 = (-1)n(2n!)/n!.n!

                 = (-1)n 1.2.3.4.5. …. (2n-1).(2n)/n!.n!

                 = (-1)n [1.3.5…. (2n-1)].2n[1.2.3…..n]/(1.2.3…n).n!

                              = (-2)n[1.3.5…(2n-1)].2n/n!
 

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