How to calculate $\sum_{i=0}^{i=x}(1+r)^{x-i}$, where $c,x$ are constant

Question

How to calculate $\sum_{i=0}^{i=x}c^{x-i}$, where $c,x$ are constant?

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My real world example is suppose we want to calculate the total retirement savings over time where $x$ denotes years between now and retirement, $x-i$ denotes years until retirement, $r$ denotes the rate of return, and $M$ is your annual contribution. If we wanted to know how much savings we have at the end, then we would do the following:

$M(1+r)^x+M(1+r)^{x-1}+M(1+r)^{x-2} + \dots + M(1+r) + M= M\sum_{i=0}^{i=x}(1+r)^{x-i}$

How do I solve this?

Answer 1

Hint: $\sum c^{x-i}=c^x\sum c^{-i}$. This series is geometric. Can you take it from here?

Answer 2

This is a partial sum of a geometric series. From here:

$$\sum_{k=0}^{n-1}ar^k = a\frac{1-r^n}{1-r}.$$

Also note that

$$\sum_{k=0}^n ar^{n-k} = \sum_{k=0}^n ar^k.$$

Substituting for your case:

$$\sum_{i=0}^x c^{x-i} = \sum_{i=0}^x c^{i} = \frac{1-c^{x+1}}{1-c}.$$

You can multiply the right side by $M$ and replace $c$ with $1+r$ to solve the application problem:

$$\sum_{i=0}^x M(1+r)^{x-i} = M\sum_{i=0}^x (1+r)^{i} = M\frac{1-(1+r)^{x+1}}{1-(1+r)}.$$