Find the sum of the series `1^2+3^2+5^2+ ton` terms.

Explanation

Let `T_(n)` be the nth term of this series, then
`T_(n)=[1+(n-1)2]^(2)=(2n-1)^(2)=4n^(2)-4n+1`
`therefore` Sum of n terms `S_(n)=sumT_(n)=4sumn^(2)-4sumn+sum1`
`=(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n`
`=(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n`
`=(n(4n^(2)-1))/(3)`