Finding Coef. of x2n: An Easy Way?

[Miike012]

After multiplication the coef. of x²ⁿ was found which is highlighted in red in the paint document.

My question is, is there an easy way of finding the coef. without actually going through the process of squaring the series given and locating the pattern for the coef. for x²ⁿ?

What I tried doing was first I found the general n^th term which is

(this is the general term after n=1)
(-1)^n-1x^n-1/(n-1)^1/4

Then I found the (n+1)^th term which is
(-1)^n-1xⁿ/(n)^1/4

Then I multiplied them together and got

x^2n-1/(n^1/4(n-1)^1/4)

and the coef. of x²ⁿ is the sum of

x^-1/(n^1/4(n-1)^1/4)

From n = 2 to n = n+1

 

[lurflurf]

See
http://en.wikipedia.org/wiki/Cauchy_product

we have
$$\left( \sum_{k=0}^\infty a_k x^k \right) \left( \sum_{l=0}^\infty b_l x^l \right)=\sum_{j=0}^\infty c_j x^j \\
\text{where} \\
c_j=\sum_{i=0}^j a_i b_{j-i}$$

provided we have enough convergence.