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UPSEE 2018 Solved Paper

Question : 130 of 150

Marks: +1, -0

\left(x+\frac{1}{x}\right)^{2n}

Solution:

Since, middle term has greatest coefficient.

\therefore

Greatest coefficient = Coefficient of middle term... (i)

Since,

2n

is an even.

\therefore

Middle term

=\left(\frac{2n}{2}+1\right)

th term

=(n+l)

th term

=T_{n+1}={}^{2n}C_{n}x^{n-n}\left(\frac{1}{x}\right)^{n}

={}^{n}C_{n}x^{n}\frac{1}{x^{n}}={}^{2n}C_{n}

Now, coefficient of middle term

={}^{2n}C_{n}=\frac{2n!}{n!n!}

\therefore

Greatest coefficient of

\left(x+\frac{1}{x}\right)^{2n}={}^{n}C_{n}[

from Eq. (i)

]

=\frac{2n!}{n!n!}

=\frac{2n(2n-1)(2n-2)(2n-3)\cdots 4\cdot 3\cdot 2\cdot 1}{n!n!}

=\frac{[1\cdot 3\cdot5\cdots (2n-1)]2^{n}[n(n-1)(n-2)\cdots 2\cdot 1]}{n!n!}

=\frac{[1\cdot 3\cdot 5\cdots(2n-1)]2^{n}\cdot n!}{n!n!}

=\frac{[1\cdot 3\cdot 5\cdots(2n-1)2^{n}]}{n!}

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