SOLUTION: What is the sum of an arithmetic series with Sigma (n on top, k=50 on the bottom, and the equation is just k.) This should be easy but I'm just having a major brain fart today...

Algebra ->  Sequences-and-series -> SOLUTION: What is the sum of an arithmetic series with Sigma (n on top, k=50 on the bottom, and the equation is just k.) This should be easy but I'm just having a major brain fart today...      Log On


   

Question 603123: What is the sum of an arithmetic series with Sigma (n on top, k=50 on the bottom, and the equation is just k.) This should be easy but I'm just having a major brain fart today...
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
S=sum%28k%2Ck=50%2Cn%29=50+51+52+ ... +(n-2)+(n-1)+n is a sum of n-49 terms.
It can also be written as:
S=sum%28k%2C1%2Cn%29-sum%28k%2C1%2C49%29=sum%28k%2C0%2Cn%29-sum%28k%2C0%2C49%29
There are many ways to approach the problem using "pre-cooked" formulas and results.
Otherwise, cooking from scratch,
2%2AS=S%2BS=50+51+52+ ... +(n-2)+(n-1)+n+50+51+52+ ... +(n-2)+(n-1)+n, with 2(n-49) terms.
Pairing terms (associative property) "head to tail" (first with last, second with second last, and so on), we get the sum of n-49 pair sums (n-49 sums of two numbersin brackets)
2S=(50+n)+(51+(n-1))+(52+(n-2)+ ...=50+50+50+ ... = 50(n-49)
So, if 2%2AS=50%28n-49%29 ---> S=highlight%2825%28n-49%29%29