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

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

A

`(n(2n-1)(2n+1))/(3)`

B

`(n(2n+1)(2n+1))/(3)`

C

`(n(2n-1)(2n-1))/(3)`

D

`(n(2n+1)(2n-1))/(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
`1^(2)+3^(2)+5^(2)+…=Sigma(2n-1)^(2)`
`=Sigma(4n^(2)-4n+1)`
`=4Sigman^(2)-4Sigman+n`
`=4(n(n+1)(2n+1))/3-2n(n+1)+n`
`=n/3(4n^(2)-1)`
`=(n(2n-1)(2n+1))/3`
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