# Finding the nth term

1. Aug 10, 2013

### 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 nth term which is

(this is the general term after n=1)
(-1)n-1xn-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

x2n-1/(n1/4(n-1)1/4)

and the coef. of x2n is the sum of

x-1/(n1/4(n-1)1/4)

From n = 2 to n = n+1

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2. Aug 10, 2013

### 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.