SOLUTION: What is the formula for: 1+(1+2)+(1+2+3+4)+(1+2+3+4+5+6+7+8)+... and how is it derived? I do know the formula btw (-: but I am not sure how to derive it.

Algebra ->  Sequences-and-series -> SOLUTION: What is the formula for: 1+(1+2)+(1+2+3+4)+(1+2+3+4+5+6+7+8)+... and how is it derived? I do know the formula btw (-: but I am not sure how to derive it.      Log On


   

Question 877695: What is the formula for:
1+(1+2)+(1+2+3+4)+(1+2+3+4+5+6+7+8)+...
and how is it derived? I do know the formula btw (-: but I am not sure how to derive it.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Recall 1+2+...+n = n(n+1)/2. If we can find a general formula for , we are essentially done.

Rewrite as . The latter term is equal to . Note that the first sum is geometric with common ratio 4 (can be written as 1^2 + 4^2 + ... + 4^n) and is equal to .

Therefore the sum is , which I'll leave you to simplify.