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I would like to proceed from where DanaJ leftI. 1 + 2 + 3 + ... + n = n(n+1)/2
II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.
Voila!! Question designed to explore the link between the two formulas.gmat740 wrote:I would like to proceed from where DanaJ leftI. 1 + 2 + 3 + ... + n = n(n+1)/2
II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.
II. [n(n+1)/2 ]*[(2n+1)/3]
= A*[(2n+1)/3]
put n= 22
A*[45/3]
=A*15 or 15A
Hope this helps
Karan
You can use that formula only if its a arithmetic progresssion.deepoe wrote:Can't u use the formula A + (n-1)D ?
