If T_r = ^{2016}C_r x^{2016-r} for r = 0, 1, | vedclass

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(D) Given $T_r = {^{2016}C_r} x^{2016-r}$.
Consider the binomial expansion: $(x + 1)^{2016} = \sum_{r=0}^{2016} {^{2016}C_r} x^{2016-r} = T_0 + T_1 +

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$(x^2 + 1)^{2016}$

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(D) Given $T_r = {^{2016}C_r} x^{2016-r}$.
Consider the binomial expansion: $(x + 1)^{2016} = \sum_{r=0}^{2016} {^{2016}C_r} x^{2016-r} = T_0 + T_1 + T_2 + \dots + T_{2016}$.
Let $x$ be replaced by $x$ and $1$ be replaced by $i$ in the expansion of $(x+i)^{2016}$:
$(x + i)^{2016} = \sum_{r=0}^{2016} {^{2016}C_r} x^{2016-r} i^r = (T_0 - T_2 + T_4 - \dots) + i(T_1 - T_3 + T_5 - \dots)$.
Let $A = (T_0 - T_2 + T_4 - \dots)$ and $B = (T_1 - T_3 + T_5 - \dots)$.
Then $A + iB = (x + i)^{2016}$.
Taking the modulus squared on both sides:
$|A + iB|^2 = |(x + i)^{2016}|^2 = |x + i|^{2 	imes 2016}$.
$A^2 + B^2 = (\sqrt{x^2 + 1^2})^{2 	imes 2016} = (x^2 + 1)^{2016}$.
Thus, $A^2 + B^2 = (x^2 + 1)^{2016}$.

If $T_r = ^{2016}C_r x^{2016-r}$ for $r = 0, 1, 2, \dots, 2016$,then $(T_0 - T_2 + T_4 - \dots + T_{2016})^2 + (T_1 - T_3 + T_5 - \dots - T_{2015})^2$ is equal to-

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