May 2, 2018

${\sum}_{k = 1}^{17} {k}^{2} = 1785$

#### Explanation:

We know that,

color(red)(sum_(i=1)^ni^2=n/6(n+1)(2n+1)

So,

${\sum}_{k = 1}^{17} {k}^{2} = {\left[\frac{n}{6} \left(n + 1\right) \left(2 n + 1\right)\right]}_{n = 17}$

${\sum}_{k = 1}^{17} {k}^{2} = \frac{17}{6} \left(17 + 1\right) \left(17 \times 2 + 1\right)$

${\sum}_{k = 1}^{17} {k}^{2} = \frac{17}{\cancel{6}} \times {\cancel{18}}^{3} \times 35$

${\sum}_{k = 1}^{17} {k}^{2} = 1785$

May 2, 2018

${\left(1\right)}^{2} + {\left(2\right)}^{2} + {\left(3\right)}^{2} + {\left(4\right)}^{2} + {\left(5\right)}^{2} + {\left(6\right)}^{2} + {\left(7\right)}^{2} + {\left(8\right)}^{2} + {\left(9\right)}^{2} + {\left(10\right)}^{2} + {\left(11\right)}^{2} + {\left(12\right)}^{2} + {\left(13\right)}^{2} + {\left(14\right)}^{2} + {\left(15\right)}^{2} + {\left(16\right)}^{2} + {\left(17\right)}^{2}$

$1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 + 169 + 196 + 225 + 256 + 289 = 1785$

May 2, 2018

See below

#### Explanation:

${\sum}_{k = 0}^{17} {k}^{2} = {0}^{2} + {1}^{2} + {2}^{2} + \ldots + {16}^{2} + {17}^{2} = 1 + 4 + 9 + . . + 256 + 289$

We can obtain this sum "manually" using a computer o calculator, but we can take advantage of the following identity

${\sum}_{k = 1}^{n} {k}^{2} = \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}$

You can obtain a demonstration at https://brilliant.org/wiki/sum-of-n-n2-or-n3/

For n=17 sum_(k=1)^17k^2=(17(17+1)(2ยท17+1))/6=1785