Question

Find
$\frac{ {1}^{2} }{1} + \frac{ {1}^{2} + {2}^{2} }{2} + \frac{ {1}^{2} + {2}^{2} + {3}^{2} }{3} + ...$
upto 'n' Terms.​

Answer 1

Step-by-step explanation:

We have,

$s _{n} = \frac{ {1}^{2} }{1} + \frac{ {1}^{2} + {2}^{2} }{2} + \frac{ {1}^{2} + {2}^{2} + {3}^{2} }{3} + ... + \frac{ {1}^{2} + {2}^{2} + {3}^{2} + .... + {n}^{2} }{n}$

$= > s_{n} = \sum _{r = 0}^{n}t _{r}$

$= >s _{n} = \sum _{r = 0}^{n} \frac{r(r + 1)(2r + 1)}{6r}$

$= > s_{n} = \sum _{r = 0}^{n} \frac{ (2 {r}^{2} + 3r + 1)}{6}$

$= > s_{n} = \sum _{r = 0}^{n}( \frac{ {r}^{2} }{3} + \frac{r}{3} + \frac{1}{6})$

$= > s_{n} = \frac{1}{3} \sum _{r = 0}^{n} {r}^{2} + \frac{1}{2} \sum _{r = 0}^{n}r + \frac{n}{6}$

$= > s_{n} = \frac{ n(n + 1)(2n + 1)}{18} + \frac{n(n + 1)}{4} + \frac{n}{6}$