The value of $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} . . . . a_{2 n}^{2}$ is

a_{n}

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2 a_{n}

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3 a_{n}

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\frac{n + 1}{2} (a_{0} + a_{1} + a_{2} . . . a_{2 n})

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Solution

The correct option is A $a_{n}$
$(1 + x)^{2 n} = 1 +^{2 n} C_{1} x +^{2 n} C_{2} x^{2} + . . . . +^{2 n} C_{2 n} x^{2 n}$
or,
$(1 + x)^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ ... $(i)$
Similarly,
$(1 - x)^{2 n} = a_{0} - a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ .... $(i i)$
As, $a_{0} = a_{2 n}$ , $a_{1} = a_{2 n - 1}, . . .$ and so on. So, (i) and (ii) can be written as:
$(1 - x)^{2 n} = a_{2 n} - a_{2 n - 1} x + a_{2 n - 2} x^{2} + . . . + a_{0} x^{2 n}$
$(1 + x)^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$

So, required answer is coefficient of $x^{2 n}$ in $(1 + x)^{2 n} . (1 - x)^{2 n} =$ coefficient of $x^{2 n}$ in $(1 - x^{2})^{2 n}$
$T_{r + 1} =^{2 n} C_{r} (- x^{2})^{r}$
$=^{2 n} C_{r} (- 1)^{r} (x)^{2 r}$
So, we need, $r = n$ . Hence, the answer is $(- 1)^{n} . a_{n}$
For $n - e v e n$ , the answer is option $A$ , i.e; $a_{n}$

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