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Given an equation of an exponential line how would I get the sum of all previous whole numbers in that line down to 0.

For example. With the equation: $y = 100 \times 1.3 ^x$

How would I create another equation to get the sum of all previous values given x?

for example, an equation that would get me this sum column.

|---x---|100 * 1.3 ^ x|--sum--|
------------------------------
|---0---|-----100-----|--100--|
|---1---|-----130-----|--230--|
|---2---|-----169-----|--399--|
|---3---|-----220-----|--619--|

I've searched and found triangle numbers (x2 + x)/2 which is almost what I'm searching for but have failed to find a way to implement this using an equation.

Jean-Claude Arbaut
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Jafer
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1 Answers1

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So you want to compute the sum

$$\sum_{k=0}^n 100 a^k$$

For $a=1.3$

Then, for any $a\neq1$,

$$\sum_{k=0}^n 100 a^k=100\sum_{k=0}^n a^k=100\frac{a^{n+1}-1}{a-1}$$

Have a look at: https://en.wikipedia.org/wiki/Geometric_series

Jean-Claude Arbaut
  • 23,277
  • Thanks for the answer, that wiki link pointed me in the right direction. I feel the ending formula was meant to be $$\sum_{k=0}^n 100 a^k=100\sum_{k=0}^n a^k=100\frac{a^{n+1}-1}{a-1}$$ – Jafer Oct 13 '16 at 06:39
  • Sorry meant to type a new line not submit $$\sum_{k=0}^n 100 a^k=100\sum_{k=0}^n a^k=100\frac{a^{n+1}-1}{a-1}$$ – Jafer Oct 13 '16 at 06:40
  • Sorr this one: $$\sum_{k=0}^n 100 a^k=100\sum_{k=0}^n a^k=100\frac{1 - a^{n+1}}{1 - a}$$ – Jafer Oct 13 '16 at 06:40
  • @Jafer Multiply by $-1$ in denominator and numerator: it's the same formula – Jean-Claude Arbaut Oct 13 '16 at 06:41