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Find the sum to 'n' terms of the series ...

Find the sum to 'n' terms of the series `1^(2)+3^(2)+5^(2)+.....+`

Text Solution

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1,3,5 are in A.P.
`n^(th)` term `=1+(n-1)2=2n-1`
`thereforen^(th)` term of the series is `T_(n)=(2n-1)^(2)=4n^(2)-4n+1`
`S_(n)=sumT_(n)=sum(4n^(2)-4n+1)=4sumn^(2)-4sumn+sum1`
`thereforeS_(n)=(4n(n+1)(2n+1))/(6)-4(n(n+1))/(2)+n=n[(2(n+1)(2n+1))/(3)-2(n+1)+1]`
`=(n)/(3)[2(2n^(2)+3n+1)-(6n+6)+3]=(n)/(3)(4n^(2)+6n+2-6n-3)=(n)/(3)(4n^(2)-1)`
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