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

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

To expand the expression [1 + x^2(1 + x)]^7 in ascending powers of x, we can use the binomial theorem. According to the binomial theorem, if we raise a binomial expression (a + b)^n to a positive integer power, the result can be expressed as the sum of the terms in the form of \(C(n, k) \cdot a^{n-k} \cdot b^k\), where \(C(n, k)\) is the binomial coefficient, given by the formula:

\(C(n, k) = \frac{n!}{k!(n-k)!}\)

In this case, we have the expression [1 + x^2(1 + x)]^7. Let's expand it:

Step 1: Replace the substituted term with a single symbol:
Let \(u = 1 + x^2(1 + x)\), so we need to expand \(u^7\).

Step 2: Apply the binomial theorem:
\(u^7 = C(7, 0) \cdot u^7 + C(7, 1) \cdot u^6 + C(7, 2) \cdot u^5 + C(7, 3) \cdot u^4 + C(7, 4) \cdot u^3 + C(7, 5) \cdot u^2 + C(7, 6) \cdot u^1 + C(7, 7) \cdot u^0\)

Step 3: Substitute the symbol back:
Substituting \(u = 1 + x^2(1 + x)\) back into the equation, we get:

\((1 + x^2(1 + x))^7 = C(7, 0) \cdot (1 + x^2(1 + x))^7 + C(7, 1) \cdot (1 + x^2(1 + x))^6 + C(7, 2) \cdot (1 + x^2(1 + x))^5 + C(7, 3) \cdot (1 + x^2(1 + x))^4 + C(7, 4) \cdot (1 + x^2(1 + x))^3 + C(7, 5) \cdot (1 + x^2(1 + x))^2 + C(7, 6) \cdot (1 + x^2(1 + x))^1 + C(7, 7) \cdot (1 + x^2(1 + x))^0\)

Step 4: Simplify each term:
Let's calculate each term in ascending powers of x:

First term:
\(C(7, 0) = \frac{7!}{0!(7-0)!} = 1\)
Simplifying further: \(1 \cdot (1 + x^2(1 + x))^7\)

Second term:
\(C(7, 1) = \frac{7!}{1!(7-1)!} = 7\)
Simplifying further: \(7 \cdot (1 + x^2(1 + x))^6\)

Third term:
\(C(7, 2) = \frac{7!}{2!(7-2)!} = 21\)
Simplifying further: \(21\cdot (1 + x^2(1 + x))^5\)

And so on...

We continue this process until we reach the term in \(x^8\).