If $(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}$ , then value of $(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}$ is

2^{10}

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2

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2^{20}

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2^{30}

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Solution

The correct option is A $2^{10}$
Here $a_{i} =^{10} C_{i}$
ie, ${(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2}$
$= {(^{10} C_{0} -^{10} C_{2} +^{10} C_{4} -^{10} C_{6} +^{10} C_{8} -^{10} C_{10})}^{2} + {(^{10} C_{1} -^{10} C_{3} +^{10} C_{5} -^{10} C_{7} +^{10} C_{9})}^{2}$
$\Rightarrow {((^{10} C_{0} -^{10} C_{10}) + (^{10} C_{8} -^{10} C_{2}) + (^{10} C_{4} -^{10} C_{6}))}^{2} + {((- 1) {(- 2)}^{\frac{10}{2}})}^{2} = 2^{10}$
Since $\sum_{i = 0}^{[n / 2]}^{n} C_{2 i} = ⎧ ⎨ ⎩ \begin{matrix} 0, If \frac{n + 2}{4} \in Integers {(- 1)}^{[(n + 2) / 4]} 2^{[n / 2]} \end{matrix}$
and $\sum_{i = 0}^{[n / 2]}^{n} C_{2 i + 1} = ⎧ ⎨ ⎩ \begin{matrix} 0, If \frac{n}{4} \in Integers {(- 1)}^{[n / 4]} 2^{[n / 2]} \end{matrix}$
Option A is correct

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