if the sum of first n terms of an ap is cn^2,then the sum of squares of these n terns isa)n(4n^2-1)c^2/6b)n(4n^2+1)c^2/3c)n(4n^2-1)c^2/3d)n(4n^2+1)c^2/6

Dear Arpit,

Saying the sum to n is one term less than the sum to (n+1)

We expand the sums:


As expected, the cubic terms cancel, and we rearrange the formula to have the sum of the squares on the left:

Expanding the cube and summing the sums:

Adding like terms:

Dividing throughout by 3 gives us the formula for the sum of the squares:

Best Of luck

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Aman Bansal

Askiitian Expert

Tn=Sn-Sn-1

therefore....do likethat..taking Sn=cn^2 and Sn-1 as c(n-1)^2;

subtract.......u will get Tn

ΣTn^2 is what we need..............so,that is=sigma of that RHS expression..evaluate it and the asnwer will come as option (c).