Q2014/107) find coefficient of a^2 in the following expression

 $(1-a^2)+(1-a^2)^2+(1-a^2)^3+(1-a^2)^4+(1-a^2)^5+(1-a^2)^6$
       $+(1-a^2)^7+(1-a^2)^8+(1-a^2)^9+(1-a^2)^{10}+(1-a^2)^{11}+(1-a^2)^{12}$

if we put $x= (1-a^2)$
we get given expression as
$x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}$= $x\dfrac{1-x^{12}}{1-x}$
= $(1-a^2)\dfrac{1-(1-a^2)^{12}}{a^2}$
= $\dfrac{(1-a^2)-(1-a^2)^{13}}{a^2}$

now $\dfrac{1-a^2}{a^2}$ shall not contribute to $a^4$ so we need to find the coefficient of $a^6$ in $-(1-a^2)^{13}$ which is ${13\choose 3}$

that is the ans