∑

_{k=1}

^{n}[A ρ

^{k}] = A ρ (ρ

^{n}-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 ρ

^{k}, 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=1}

^{n}[2 (1/3)

^{k}] = 2/3 + 2/9 + 2/27 + 2/81 + ... + 2/3

^{n}.

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.

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