If $T_r=^{2016}C_rx^{2016-r}$, 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 - ;

Let $i$ represent iota.

$(T_0-T_2+T_4+\dots+T_{2016}) = (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})\dots(1)$

Now, multiply $(T_1-T_3+T_5+\dots-T_{2015})$ with $-i$ to get $-(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})\dots(2)$

$\implies (T_0-T_2+T_4+\dots+T_{2016})^2+(T_1-T_3+T_5+\dots-T_{2015})^2$

$= (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 + i^4(T_1-T_3+T_5+\dots-T_{2015})^2$

$= (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 -(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})^2$

$= (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$ (As $a^2-b^2 = (a+b)(a-b)$)

Now, consider these terms seperately

$(T_0+iT_1+i^2T_2+\dots)=(^{2016}C_0x^{2016}i^0 +^{2016}C_1x^{2015}i^1\dots) = (i+x)^{2016}$

$(T_0-iT_1+i^2T_2-i^3T_3\dots) = (^{2016}C_0(ix)^{2016} +^{2016}C_1(ix)^{2015}\dots) = (1+ix)^{2016}$

$\therefore (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$

$= (i+x)^{2016}\times(1+ix)^{2016}$

$= [(i+x)(1+ix)]^{2016}$

$= [i-x+x+ix^2]^{2016}$

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