If xp occurs in the equation x 2 - 1 x 2n- prove that its coefficient is - 2n - 1 3 4n-p - 1 3 2n-p-

#10; #10; ⟹ General term: #10; T _ r+1 =^ 2n C _ r ( x ^ 2 ) #160; ^ 2n r #10;( 1 x #160; ) #160; ^ r #10; #10; =^ 2n C ...

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Binomial Coefficients

If xp occur...

Question

If $x^{p}$ occurs in the equation ${(x^{2} + \frac{1}{x})}^{2 n}$ ., prove that its coefficient is $\frac{| 2 n - - -}{| \frac{1}{3} (4 n - p) - --------- - | \frac{1}{3} (2 n + p) - --------- -}$ .

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Solution

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x^{p}

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

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

x^{p}

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

\frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!}

x^{P}

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

\frac{1}{1 + x} + \frac{2}{1 + x^{2}} + \frac{4}{1 + x^{4}} + . . . + \frac{2^{n}}{1 + x^{2^{n}}}

= \frac{1}{1 -} \frac{2^{n + i}}{1 + x^{2^{n + i}}}

\frac{1}{x + x^{p - q}} + \frac{1}{1 + x^{q / - p}} = 1

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