Question

Find \[\displaystyle\sum_{r=1}^{n}\frac{1^2 + 2^2 + 3^2+...+r^2}{2r + 1}\]

Answer 1

We know that

1² + 2² + 3² + ... + n² = `("n"("n" + 1)(2"n" + 1))/6`

∴ 1² + 2² + 3² + ... + r² = `("r"("r" + 1)(2"r" + 1))/6`

∴ `(1^2 + 2^2 + 3^2 + ...  "r"^2)/(2"r" + 1) = ("r"("r" - 1))/6`

∴ \[\displaystyle\sum_{r=1}^{n}\frac{1^2 + 2^2 + 3^2+...+r^2}{2r + 1}\]

= \[\displaystyle\sum_{r=1}^{n}\frac{r(r+1)}{6} = \frac{1}{6}\displaystyle\sum_{r=1}^{n}(r^2 + r)\]

= \[\frac{1}{6}{(\displaystyle\sum_{r=1}^{n} r^2+ \displaystyle\sum_{r=1}^{n} r)}\]

= `1/6[("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2]`

= `1/6 xx ("n"("n" + 1))/2 ((2"n" + 1)/3 + 1)`

= `("n"("n" + 1))/12 ((2"n" + 1 + 3)/3)`

= `("n"("n" + 1)(2"n" + 4))/36`

= `(2"n"("n" + 1)("n" + 2))/36`

= `("n"("n" + 1)("n" + 2))/18`