A Alphlax New member Joined Sep 10, 2013 Messages 9 Sep 10, 2013 #1 So I'm stumped on this question 1+4+9+16+...+100=? I know in between is +3+5+7 but is it like 2n(99)?
So I'm stumped on this question 1+4+9+16+...+100=? I know in between is +3+5+7 but is it like 2n(99)?
D daon2 Full Member Joined Aug 17, 2011 Messages 999 Sep 10, 2013 #2 \(\displaystyle \displaystyle \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Sep 11, 2013 #3 Hello, Alphlax! 1 + 4 + 9 + 16 + ... + 100 = ? I know in between is +3+5+7 but is it like 2n(99)? What does that mean? Click to expand... You are asked to add: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Hello, Alphlax! 1 + 4 + 9 + 16 + ... + 100 = ? I know in between is +3+5+7 but is it like 2n(99)? What does that mean? Click to expand... You are asked to add: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
A Alphlax New member Joined Sep 10, 2013 Messages 9 Sep 11, 2013 #5 daon2 said: \(\displaystyle \displaystyle \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\) Click to expand... Why is it something over six?
daon2 said: \(\displaystyle \displaystyle \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\) Click to expand... Why is it something over six?
S srmichael Full Member Joined Oct 25, 2011 Messages 848 Sep 11, 2013 #6 Alphlax said: Why is it something over six? Click to expand... Because that is what the formula is.