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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 terns is a)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

if the sum of first n terms of an ap is cn^2,then the sum of squares of these n terns is 


a)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

Grade:11

2 Answers

Aman Bansal
592 Points
10 years ago

Dear Arpit,

sumSquares1.gif
Saying the sum to n is one term less than the sum to (n+1)

We expand the sums:
sumSquares2.gif

As expected, the cubic terms cancel, and we rearrange the formula to have the sum of the squares on the left:
sumSquares3.gif
Expanding the cube and summing the sums:
sumSquares4.gif
Adding like terms:
sumSquares5.gif

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

Best Of luck

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Arjun Chaudhuri
18 Points
10 years ago

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).

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