Find the sum to n terms of the series, whose `n^(t h)`terms is given by : `(2n-1)^2`

Answer

N/a

Explanation

`because` nth term of the series `= (2n - 1)^(2)`
`= 4n^(2) - 4n + 1`
`therefore` Sum of n terms of the series
`S_(n) = sum(4n^(2) - 4n + 1)`
`= 4 sum n^(2) - 4 sum n + n`
`= 4 * (n(n + 1)(2n + 1))/(6) - (4n(n + 1))/(2) + n`
`= (4n(n + 1))/(2) ((2n + 1)/(3) - 1) + n`
`= 2n(n + 1) ((2n - 2))/(3) + n`
`=(4)/(3)n (n + 1) (n - 1) + n`
`=(n)/(3)[4(n + 1)(n - 1) + 3] = (n)/(3)[4(n^(2) - 1) + 3]`
`= (n)/(3)(4n^(2) - 1) = (n)/(3)(2n + 1)(2n - 1)`
Therefore, sum of n terms of series
`= (n)/(3) (2n + 1)(2n -1)`