If I have an x where x starts at x=5, and each step adds 10, so that x1=5, x2=15, x3=25, etc...so that if there were 3 steps the answer would be 5+15+25=45.
This is most properly answered by https://www.quora.com/What-is-the-sum-of-1-to-50/answer/Arunabh-Kashyap-1 , in this case the answer is S=n/2{2a+(n−1)d where n=number of steps, d=amount increased, and a is the start point, so:
100/2*(2*5+(100-1)*10)=50000
I was taught it as a shortened formula of average of first number and last number (5 and 995, for an average of 500) times the number of steps. (500*100=50000)
Now, that works for a set additive expansion, but if I use a multiplier instead of an addition, I find it stops working.
If I start at x=5 and multiply each step by 2, so that x1=5, x2=10, x3=20, at x3 the sum is 5+10+20=35
At 100 steps the actual answer to 3 significant digits is 6.34E+30. Using my shortened formula above, I get 1.58E+32. I have no idea how to modify the original formula.
How can I modify this as a general formula to figure out large stepped problems like this with generic starting points and for various multiplicative values? Also, if I need to figure out the value for the last step, how can I do that without using a spreadsheet? :)
It's called a geometric series, and there's a well-known general summation formula:
$$\sum_{k=0}^n ar^k = a(\frac{1-r^{n+1}}{1-r})$$
So if you have a starting value $a=5$, and multiplier $r=2$, the sum of the first 100 terms ($n=99$) is $5\frac{1-2^{100}}{1-2} \approx 6.338253 \times 10^{30} $