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Algebra Level 4

For a certain constant c c , there exists an arithmetic progression such that the sum of the first n n terms is equal to c n 2 c n^2 . What is the sum of squares of the first n n terms?

n ( 4 n 2 − 1 ) c 2 3 \frac { n(4{ n }^{ 2 }-1){ c }^{ 2 } }{ 3 } n ( 4 n 2 + 1 ) c 2 3 \frac { n(4{ n }^{ 2 }+1){ c }^{ 2 } }{ 3 } n ( 4 n 2 + 1 ) c 2 6 \frac { n(4{ n }^{ 2 }+1){ c }^{ 2 } }{ 6 } n ( 4 n 2 − 1 ) c 2 6 \frac { n(4{ n }^{ 2 }-1){ c }^{ 2 } }{ 6 }

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Manish Dash by Manish Dash , 22, India

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1 solution

Given the sum of the AP is c n 2 cn^2 , it can be verified using the formula for the sum of AP and comparing powers of n that a=c and d=2c.

Now, the kth term of the AP will be T k = 2 c k − k = c ( 2 k − 1 ) T_k=2ck-k=c(2k-1) . Its square is T k 2 = c 2 ( 4 k 2 − 4 k + 1 ) T_k^2=c^2(4k^2-4k+1) . The sum we desire is S n = Σ k = 1 k = n c 2 ( 4 k 2 − 4 k + 1 ) S_n=\Sigma_{k=1}^{k=n}c^2(4k^2-4k+1)

which is simplified to the given answer after some algebra. An easy picking!

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