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Question
Prove that the term independent of x in the expansion of \[\left( x + \frac{1}{x} \right)^{2n}\] is \[\frac{1 \cdot 3 \cdot 5 . . . \left( 2n - 1 \right)}{n!} . 2^n .\]
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Solution
\[\text{ Given:} \]
\[ \left( x + \frac{1}{x} \right)^{2n} \]
\[\text{ Suppose the term independent of x is the (r + 1) th term } . \]
\[ \therefore T_{r + 1} =^{2n}{}{C}_r x^{2n - r} \frac{1}{x^r}\]
\[ = ^{2n}{}{C}_r x^{2n - 2r} \]
\[\text{ For this term to be independent of x, we must have: } \]
\[2n - 2r = 0\]
\[ \Rightarrow n = r\]
\[ \therefore \text{ Required coefficient } =^ {2n}{}{C}_n \]
\[ = \frac{(2n)!}{(n! )^2}\]
\[ = \frac{\left\{ 1 \cdot 3 \cdot 5 . . . \left( 2n - 3 \right)\left( 2n - 1 \right) \right\}\left\{ 2 \cdot 4 \cdot 6 . . . \left( 2n - 2 \right)\left( 2n \right) \right\}}{(n! )^2}\]
\[ = \frac{\left\{ 1 \cdot 3 \cdot 5 . . . \left( 2n - 3 \right)\left( 2n - 1 \right) \right\} 2^n}{n!}\]
\[\]
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