NIMCET 2019 — Mathematics PYQ

Calculation:

Given, $(1+x-2x^2{)}^6=1+a_{1}x+a_2{x}^2+...+a_{12}{x}^{12}$

Put, $x=1$,
$(1+1-2\times 1^2{)}^6=1+a_1+a_2+...+a_{12}$
$1+a_1+a_2+...+a_{12}=0$ ... (1)

Now, put $x=-1$
$(1-1-2\times (-1{)}^2{)}^6=1-a_1+a_2-...+a_{12}$
$(2{)}^6=1-a_1+a_2-...+a_{12}$ ... (2)

Add (1) and (2)

$2(1+a_2+a_4+...a_{12})=0+2^6$

$1+a_2+a_4+...a_{12}=32$

$\therefore a_2+a_4+...a_{12}=31$