If $T_{r} =^{2016} C_{r} x^{2016 - r}$ , for $r = 0, 1,, . . . .2016$ , then $(T_{0} - T_{2} + T_{4} . . . . + T_{2016})^{2} + (T_{1} - T_{3} + T_{5} . . . . T_{2015})^{2}$ is equal to -

(X^{2} - 1)^{1008}

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(X + 1)^{2016}

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(X^{2} - 1)^{2016}

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(X^{2} + 1)^{2016}

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Solution

The correct option is B $(X^{2} + 1)^{2016}$
Let $i$ represent iota.
$(T_{0} - T_{2} + T_{4} + \dots + T_{2016}) = (T_{0} + i^{2} T_{2} + i^{4} T_{4} + \dots + i^{2016} T_{2016}) \dots (1)$
Now, multiply $(T_{1} - T_{3} + T_{5} + \dots - T_{2015})$ with $- i$ to get $- (i T_{1} + i^{3} T_{3} + i^{3} T_{5} + \dots + i^{2015} T_{2015}) \dots (2)$
$⟹ (T_{0} - T_{2} + T_{4} + \dots + T_{2016})^{2} + (T_{1} - T_{3} + T_{5} + \dots - T_{2015})^{2}$
$= (T_{0} + i^{2} T_{2} + i^{4} T_{4} + \dots + i^{2016} T_{2016})^{2} + i^{4} (T_{1} - T_{3} + T_{5} + \dots - T_{2015})^{2}$
$= (T_{0} + i^{2} T_{2} + i^{4} T_{4} + \dots + i^{2016} T_{2016})^{2} - (i T_{1} + i^{3} T_{3} + i^{3} T_{5} + \dots + i^{2015} T_{2015})^{2}$
$= (T_{0} + i T_{1} + i^{2} T_{2} + \dots + i^{2016} T_{2016}) (T_{0} - i T_{1} + i^{2} T_{2} - i^{3} T_{3} \dots)$ (As $a^{2} - b^{2} = (a + b) (a - b)$ )
Now, consider these terms seperately
$(T_{0} + i T_{1} + i^{2} T_{2} + \dots) = (^{2016} C_{0} x^{2016} i^{0} +^{2016} C_{1} x^{2015} i^{1} \dots) = (i + x)^{2016}$
$(T_{0} - i T_{1} + i^{2} T_{2} - i^{3} T_{3} \dots) = (^{2016} C_{0} (i x)^{2016} +^{2016} C_{1} (i x)^{2015} \dots) = (1 + i x)^{2016}$
$∴ (T_{0} + i T_{1} + i^{2} T_{2} + \dots + i^{2016} T_{2016}) (T_{0} - i T_{1} + i^{2} T_{2} - i^{3} T_{3} \dots)$
$= (i + x)^{2016} \times (1 + i x)^{2016}$
$= [(i + x) (1 + i x)]^{2016}$
$= [i - x + x + i x^{2}]^{2016}$
$= (1 + x^{2})^{2016}$

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