The second problem on the project introduces a new closed form for a sum:
∑k=1n [A ρk] = A ρ (ρn-1)/(ρ-1).
Unfortunately, too many of you are still intimidated simply by the symbols that are used.
The formula for the geometric sequence, A ρk, is like an exponential, except the power is an integer variable rather than a continuous variable like x. For example, if A=2 and ρ=1/3, we have terms that are increasing powers of (1/3) times 2:
2/3 (k=1), 2/9 (k=2), 2/27 (k=3), 2/81 (k=4), etc.
The summation is simply the sum of these values:
∑k=1n [2 (1/3)k] = 2/3 + 2/9 + 2/27 + 2/81 + ... + 2/3n.
The closed form gives a formula answering the value of this sum:
2(1/3)[(1/3)n-1]/[(1/3) - 1] = (2/3)*[(1/3)n-1]/(-2/3) = 1 - (1/3)n.
So when you write down the Riemann sum for the integral in question, you need to look at using the properties of exponentials so that the Riemann sum looks just like a sum of a geometric sequence. You should identify a factor that does not involve k, and this is A. You should identify the other factor as some number raised to the k power. Then you can use the closed form.