InterviewSolution
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InterviewSolution
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If the sum of first n terms of an AP is `cn^(2)`, then the sum of squares of these n terms isA. `(n(4n^(2)-1)c^(2))/(6)`B. `(n(4n^(2)+1)c^(2))/(3)`C. `(n(4n^(2)-1)c^(2))/(3)`D. `(n(4n^(2)+1)c^(2))/(6)` |
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Answer» Correct Answer - C `:.S_(n)=cn^(2)` `:. T_(n)=S_(n)-S_(n-1)=c(2n-1)` `sumt_(n)^(2)=c^(2)sum(2n-1)^(2)` `=c^(2)sum(4n^(2)-4n+1)=c^(2){4sumn^(2)-4sumn+sum1}` `=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}` `=c^(2)n{(2)/(3)(2n^(2)3n+1)-2n-2+1}` `=(c^(2)n)/(3)(4n^(2)-1)=(n(4n^(2)-1)c^(2))/(3)`. |
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