Find the sum -nr-112-22-32-r2rr-1 -

Given, ∑_r=1^n1^2+2^2+...+r^2r(r+1) =∑_r=1^nr(r+1)(2r+1)/6r(r+1) =∑_r=1^n2r+16 =∑ _r=1^nr3+∑ _r=1^n16 =16n(n+1)+16n =n^2+2n6 ...

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Sum of Infinite Terms of a GP

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Question

Find the sum $\sum_{r = 1}^{n} \frac{1^{2} + 2^{2} + 3^{2} + . . . . . . . + r^{2}}{r (r + 1)}$ ?

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Solution

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t_{r} = \frac{1^{2} + 2^{2} + 3^{2} + . . . + r^{2}}{1^{3} + 2^{3} + . . . + r^{3}}

S_{n} = n \sum r = 1 {(- 1)}^{r} . t_{r},

lim n \to \infty S_{n} = \frac{2}{3}

1^{2} + 2^{2} + 3^{2} + . . . + r^{2} = \frac{r (r + 1) (2 r + 1)}{6}

1^{3} + 2^{3} + 3^{3} + . . . + r^{3} = {(\frac{r (r + 1)}{2})}^{2}

1^{2} = \frac{1}{6} [1 \times (1 + 1) \times (2 \times 1 + 1)] 1^{2} + 2^{2} = \frac{1}{6} [2 \times (2 + 1) \times (2 \times 2 + 1)] 1^{2} + 2^{2} + 3^{2} = \frac{1}{6} [3 \times (3 + 1) \times (2 \times 3 + 1)] 1^{2} + 2^{2} + 3^{2} + 4^{2} = \frac{1}{6} [4 \times (4 + 1) \times (2 \times 4 + 1)]

\sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + \sqrt{1 + \frac{1}{3^{2}} + \frac{1}{4^{2}}} + . . . . n terms

\sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . . . . + \sqrt{1 + \frac{1}{2007^{2}} + \frac{1}{2008^{2}}}

\sum_{r = 1}^{n} r = 55, find \sum_{r = 1}^{n} r^{3}

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