We will expand (1+x−x2)6 and find the coefficients of x4 and x6. Multimonial expansion states that ‌(a1+a2+a3)n=∑‌
n!
k1!k2!k3!
‌a1ka2k2a3k3 where k1+k2+k3=n In our case, a1=1,a2=x,a3=−x2 and n=6 From k1+k2+k3=6 ⇒‌‌k1=6−k2−k3 To get x4, we can have different combinations of x and x2 k1 : number of times 1 is chosen k2 : number of times x is chosen k3 : number of times x2 is chosen The equation is: k2+2k3=4 Now, if k3=0, then k2=4 and k1=2 If k3=1, then k2=2 and k1=3 If k3=2, then, k2=0,k1=4 Now, find the coefficients for each valid combination. Coefficient of ‌x4=‌
6!
2!4!0!
−‌
6!
3!2!1!
+‌
6!
4!0!2!
‌=15−60+15=−30 Now, for x6,k2+2k3=6 and k1+k2+k3=6 If k3=0, then k2=6, and k1=0 If k3=1, then k2=4, and k1=1 If k3=2 then k2=2 and k1=2 If k3=3, then k2=0, and k1=3 So, coefficient of x6=‌
6!
0!6!0!
−‌
6!
1!4!1!
+‌
6!
2!2!2!
−‌
6!
3!0!3!
=1−30+90−20=41 ∴ Sum of the coefficient of x4 and x6=41−30=11