Struggling with a geometric series problem

Question

The question is sum of ((7/6)^k) - k. The initial value is k = 20 and it goes to 40. The way I tried to do it was to separate it as sum of (7/6)^k - sum of k. Knowing that the first term in the series is (7/6)^20, I used the the formula to solve both summations: (((7/6)^20)*(1 - (7/6)^40))/(1 - (7/6)) for the first summation and 40(40+1)/2 and then I subtracted the results, but that gave me the wrong answer. I know the answer is supposed to be approximately 2 573.0865.

How do I solve this problem?

Answer 1

You have the right idea but you've made some mistakes. Writing out the terms of the sum will help you detect those mistakes. For convenience, let $a = 7/6$. Then your sum looks like this:

$$S = (a^{20} - 20) + (a^{21} - 21) + (a^{22} - 22) + \cdots + (a^{40} - 40).$$

Your first step was to rearrange the geometric parts and the arithmetic parts:

$$S = (a^{20} + a^{21} + \cdots + a^{40}) - (20 + 21 + \cdots + 40).$$

So far, so good. Your next idea was to factor out $a^{20}$:

$$S = a^{20}(1 + a + a^2 + \cdots + a^{\color{red}{20}}) - (20 + 21 + \cdots + 40).$$

The last exponent in the geometric part, is where you made your mistake, because when you now apply the formula for a finite geometric series, you should have gotten

$$S = a^{20} \frac{1 - a^{\color{red}{21}}}{1 - a} - (20 + 21 + \cdots + 40).$$

As for the arithmetic part, that is just

$$20 + 21 + \cdots + 40 = \frac{40(41)}{2} - \frac{19(20)}{2}.$$

So your sum has value

$$S = \left(\frac{7}{6}\right)^{20} \frac{(1 - (7/6)^{21})}{1 - 7/6} - \frac{40(41) - 19(20)}{2}.$$