Question

Help please? evaluate seires

Answer 1

`sum_(k=1)^17k^2=1785`

Explanation:

We know that,

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

So,

`sum_(k=1)^17k^2=[n/6(n+1)(2n+1)]_(n=17)`

`sum_(k=1)^17k^2=17/6(17+1)(17xx2+1)`

`sum_(k=1)^17k^2=17/cancel6xxcancel18^3xx35`

`sum_(k=1)^17k^2=1785`

Answer 2

`(1)^2 + (2)^2 + (3)^2 + (4)^2 + (5)^2 + (6)^2 + (7)^2 + (8)^2 + (9)^2 + (10)^2 + (11)^2 + (12)^2 + (13)^2 + (14)^2 + (15)^2 + (16)^2 + (17)^2`

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

Answer 3

See below

Explanation:

`sum_(k=0)^17k^2=0^2+1^2+2^2+...+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)^nk^2=(n(n+1)(2n+1))/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)(2·17+1))/6=1785`