Posted by **Ethan** on Sunday, October 2, 2011 at 9:04am.

Expand [1+x^2(1+x)]^7 in ascending powers of x as the term in x^8.

- Math -
**Damon**, Sunday, October 2, 2011 at 10:33am
Use binomial expansion over and over again

- Math -
**Reiny**, Sunday, October 2, 2011 at 10:40am
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

35(x^2(1+x))^3

= 35x^6( 1 + 3x + 3x^2 + x^3)

and the x^8 term from that is **105x^8**

35(x^2(1+x))^4

= 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

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