1. Find a power series for F(x)= 3/4x^3-5, where c=1 2. power series = 1/a-r 3. What I did was take a derivative to get a similar function that was easier to solve. I used 1/x^3-1. Then I found a series for that function. Which I got [tex]\sum[/tex](x^3)^n. Then I added back from my orginal problem and got [tex]\sum[/tex](4/3x)^3n does this look at all right?