If ${(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{10} x^{10}$ , then the 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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None of these

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Solution

The correct option is A $2^{10}$
We have given that
${(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}$

Put $x = i$
${(1 + i)}_{^{10}} = (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10}) + i (a_{1} - a_{3} + a_{5} - a_{7} + a_{9}) - - - - (1)$

Put $x = - i$
${(1 - i)}_{^{10}} = (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10}) + i (a_{1} - a_{3} + a_{5} - a_{7} + a_{9}) - - - - (2)$

Now on adding $(1) + (2)$ we get
$(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10}) = {(1 + i)}^{10} + {(1 - i)}^{10}$
$= {(\sqrt{2} e^{\frac{i π}{4}})}^{10} + {(\sqrt{2} e^{\frac{- i π}{4}})}^{10}$

$= 2^{5} e^{\frac{i 5 π}{4}} + 2^{5} e^{\frac{- 5 π}{4}}$

$= 2^{5} (cos \frac{π}{2} + i sin \frac{5 π}{2}) + 2^{5} [cos (\frac{- 5 π}{2}) + i sin (\frac{- 5 π}{2})]$
$= 2^{5} (0 + i) + 2^{5} (0 - i) = 0$

$a_{0} - a_{2} + a_{4} - a_{6} - a_{10} = - a_{8}$
so, ${(a_{0} - a_{2} + a_{4} - a_{6} - a_{10})}^{2} = {(a_{8})}^{2}$

Now by subtracting $(1) - (2)$ we get
$2 i (a_{1} - a_{3} + a_{5} - a_{7} + a_{9}) = 2^{5} (0 + i) - 2^{5} (0 - i)$

$= 2^{5} i + 2^{5} i = 2^{6} i$

$\Rightarrow a_{1} - a_{3} + a_{5} - a_{7} + a_{9} = 2^{5}$

$\Rightarrow {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2} = 2^{10}$
So,
${(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2} = a_{8}^{2} + 2^{10}$
$= 45^{2} + 2^{10}$
Hence, the option $(A)$ is the correct answers.

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Similar questions

{(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . a_{10} x^{10}

{(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2}

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

I f (1 + x)^{10} = a_{0} + a_{1} x + . . . . a_{10} x^{10}, t h e n (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

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(a^{0} + a^{3} + a^{6} + . . . . . . . . .) = (a^{1} + a^{4} + a^{7}) = (a^{2} + a^{5} + a^{8}) = 3^{(} n - 1),

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