posted by Ethan on .
Expand [1+x^2(1+x)]^7 in ascending powers of x as the term in x^8.
Use binomial expansion over and over again
let u = x^2(1+x)
then [1 + x^2(1+x)]^7 = (1+u)^7
= 1 + 7u + 21u^2 + 35u^3 + 35u^4 + 21u^5 + 7u^6 + u^7
= 1 + 7(x^2(1+x)) + 21(x^2(1+x))2 + 35(x^2(1+x))^3 + 35(x^2(1+x))^4 ... + (x^2(1+x))^7
to carry on seems like a rather unreasonable question, but I will assume you only want the term containing x^8
if so, then x^8 can only arise in the expansions of
35(x^2(1+x))^3 and 35(x^2(1+x))^4
= 35x^6( 1 + 3x + 3x^2 + x^3)
and the x^8 term from that is 105x^8
= 35x^8( 1 + 4x + ... + x^4)
and the only term with x^8 from that is 35x^8
so the term containing x^8 is 105x^8+35x^8 = 140x^8
check my arithmetic