Question

Find the sum of the series?

`1^2 - 2^2+3^2-4^2+5^2-6^2+... +29^2 -30^2`

Answer 1

`S=-465`

Explanation:

We know that,

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

We take,

`S=1^2-2^2+3^2-4^2+5^2-6^2+...+29^2-30^2`

`=[1^2+3^2+5^2+...+29^2]-[2^2+4^2+6^2...+30^2]`

`={1^2+color(red)(2^2)+3^2+color(red)(4^2)+5^2+color(red)(6^2)+...+29^2+color(red)(30^2)color(red)(-2^2-4^2-6^2-...-30^2)}-[2^2+4^2+6^2+...+30^2]`

`={(1^2+2^2+3^2+4^2+5^2+6^2+...+29^2+30^2)-(2^2+4^2+6^2+...+30^2)}-[2^2+4^2+6^2+...+30^2]`

`=(1^2+2^2+3^2+4^2+5^2+6^2+...+29^2+30^2)-2[2^2+4^2+6^2+...+30^2]`

`=sum_(k=1)^30 k^2-2xx2^2[1^2+2^2+3^2+...+15^2]`

`=[n/6(n+1)(2n+1)]_(n=30) color(white)(s_1)-8sum_(k=1)^15 k^2`

`=30/6(30+1)(60+1)-8[n/6(n+1)(2n+1)]_(n=15`

`={5xx31xx61}-{cancel8^4xxcancel15^5/cancel6^3(15+1)(30+1)}`

`=9455-(4xx5xx16xx31)`

`=9455-9920`

`=-465`

Answer 2

`1^2-2^2+3^2-4^2+5^2-6^2+...+29^2-30^2 = -465`

Explanation:

Note that:

`(2n-1)^2-(2n)^2 = 4n^2-4n+1-4n^2 = -4n+1`

Also:

`sum_(n=1)^N n = 1/2 N (N+1)`

being the sum of `N` terms with average value `(N+1)/2`.

So:

`1^2-2^2+3^2-4^2+5^2-6^2+...+29^2-30^2`

`=sum_(n=1)^15 ((2n-1)^2-(2n)^2)`

`=sum_(n=1)^15 (-4n+1)`

`=15-4 sum_(n=1)^15 n`

`=15-4 (1/2)(15)(15+1)`

`=15-480`

`=-465`