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

2^{10}

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2

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2^{20}

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N o n e o f t h e s e

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Solution

The correct option is A $2^{10}$
$(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . + a_{10} x^{10}$
$\Rightarrow (1 + i)^{10} = (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10}) + i (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})$
Apply mod both side.
${| 1 + i |}^{10} = \sqrt{(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}}$
So, $(a_{0} - a_{2} + a_{4} + a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}$
$= ({\sqrt{2}}^{10})^{2}$
$= 2^{10 \times 2 / 2}$
$= 2^{10}$

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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}

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}

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}

(1 + x + x^{2})^{n}

a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{r} x^{r} + . . . . . + a_{2 n} x^{2 n}

a_{0} + a_{3} + a_{6} + . . . . . = a_{1} + a_{4} + a_{7} + . . . . . = a_{2} + a_{5} + a_{8} + . . . . = 3^{n - 1}

(1 + x + x^{2}) n = a 0 + a^{1} x + a^{2} x + a^{3} x^{3} + . . . . . . . a (2^{n}) x (2^{n})

(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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