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Question

Consider $(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}$ , where $a_{0}, a_{1}, a_{2}, . . ., a_{4 n}$ are real numbers and $n$ is a positive integer. The value of $a_{2}$ is

A

^{4 n + 1} C_{2}

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B

^{3 n + 1} C_{2}

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C

^{2 n + 1} C_{2}

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D

^{n + 1} C_{2}

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Solution

The correct option is B $^{2 n + 1} C_{2}$
We have $(1 + x + x^{2})^{2 n}$ $= a_{0} + a_{1} x + a_{2} x^{2} . . . a_{4 n} x^{2 n}$ -----(i)
Hence $a_{2}$ is the coefficient of $x^{2}$
$(1 + (x + x^{2}))^{2 n}$
$= 1 +^{2 n} C_{1} (x + x^{2}) +^{2 n} C_{2} (x + x^{2})^{2}$ ------(ii)
Hence, coefficient of $x^{2}$ is
$^{2 n} C_{1} +^{2 n} C_{2}$
$= 2 n + (\frac{2 n (2 n - 1)}{2!})$
$= 2 n (\frac{2 + 2 n - 1}{2})$
$= (\frac{(2 n + 1) (2 n)}{2})$
$=^{2 n + 1} C_{2}$

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