Uhmm, I think it can also be done with geometric series:HallsofIvy said:You posted this in the "precalculus section" so I assume you wouldn't be able to use "Taylor Series". That's the only way I know to show it.
I thought about that but then integrating is not "pre-calculus" either!VietDao29 said:Uhmm, I think it can also be done with geometric series:
[tex]\sum_{i = 0} ^ \infty (a r ^ i) = \frac{a}{1 - r}, \quad |r| < 1[/tex].
So if your final sum is:
[tex]\frac{1}{1 - k}[/tex], what should your geometric series look like? Then integrating that should give you the result you want (remember to choose the appropriate C, i.e the constant of integration).