Binomial Theorem Expansion

$(1+x+ax^2)^{10}=(1+x)^10+10(1+x)^9ax^2+45(1+x)^8(ax^2)^2+...$ In which $x^2$ has coefficients respectively: $C^{10}_{4}, 10C^9_2, 45a^2$ $K=10*9*8*7/4!+10*a*9*8/2!+45*a^2+360a$ $K=f(a)=45a^2+360a+210$ This function has minimum at: $a=\boxed{-4}$

coeffficient of n th term = (10!/r!s!t!) (1^r ) (x^s ) * ((ax^2)^t).........(i)

=> x^(s+2t) = x^4 therefore s + 2*t=4 which has 3 solutions of (s,t) i.e (4,0),(2,1) &(0,2)....

Substituting in (i) we get the coefficients as 210 ,360a & 45 a^2

K=210 + 360 a + 45 a^2

dK/da =0 =>90a+360 =0 =>a= -4