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- Thread starter egal
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[math]\frac{1}{n^3}\cdot\frac{2n^3+3n^2+n}{6}=\frac{2n^3+3n^2+n}{6n^3}=\frac{\frac{2n^3}{n^3}+\frac{3n^2}{n^3}+\frac{n}{n^3}}{6}=\frac{2+\frac{3}{n}+\frac{1}{n^2}}{6}[/math]

[math]\frac{1}{n^3}\cdot\frac{2n^3+3n^2+n}{6}=\frac{2n^3+3n^2+n}{6n^3}=\frac{\frac{2n^3}{n^3}+\frac{3n^2}{n^3}+\frac{n}{n^3}}{6}=\frac{2+\frac{3}{n}+\frac{1}{n^2}}{6}[/math]

okayyy, I read about sum of consecutive squares formula and got it, Thanks a lot!The sum of consecutive squares formula. Then we divided the 3 terms in the numerator by n^3. There is a typo: the last 2 should be 1.

There you are! These formulas are pretty useful:okayyy, I read about sum of consecutive squares formula and got it, Thanks a lot!

[math]\sum_{i=1}^n{i}=\frac{n(n+1)}{2}\\\\\sum_{i=1}^n{i^2}=\frac{n(n+1)(2n+1)}{6}\\\\\sum_{i=1}^n{i^3}=\frac{n^2(n+1)^2}{4}\\\\\text{P.S. You can prove them by induction ;)}[/math]

Have a nice day!

thanks a lot again, you too!There you are! These formulas are pretty useful:

[math]\sum_{i=1}^n{i}=\frac{n(n+1)}{2}\\\\\sum_{i=1}^n{i^2}=\frac{n(n+1)(2n+1)}{6}\\\\\sum_{i=1}^n{i^3}=\frac{n^2(n+1)^2}{4}\\\\\text{P.S. You can prove them by induction ;)}[/math]

Have a nice day!