Let $n \in N$ , such that $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . a_{2 n} x^{2 n}$ The value of $a_{r}$ when $(0 \leq r \leq 2 n)$

a_{2 n - r}

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a_{n - r}

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a_{2 n}

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n . a_{2 n - 1}

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Solution

The correct option is A $a_{2 n - r}$
We have
${(1 + x + x^{2})}^{n}$ $=$ $\sum_{r = 0}^{2 n} a_{r} x^{r}$
Replace $x$ by $\frac{1}{x}$
Therefore, $(1 + \frac{1}{x} + \frac{1}{x^{2}}) =$ $\sum_{r = 0}^{2 n} a_{r} x^{r}$
$\Rightarrow$ ${(1 + x + x^{2})}^{n}$ $=$ $\sum_{r = 0}^{2 n} a_{r} x^{2 n - r}$
$\Rightarrow$ $\sum_{r = 0}^{2 n} a_{r} x^{r}$ $=$ $\sum_{r = 0}^{2 n} a_{r} x^{2 n - r}$
So, equating both the sides, we get $a_{r}$ $=$ $a_{2 n - r}$ .

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