If the sum of first n terms of an AP is $$cn2$$, then the sum of squares of these n terms is $$n4n2$$−$$1c2k$$ then k is

$$Sn=cn2Sn$$−$$1=c(n$$−$$1)2=cn2+c$$−$$2cnTn=2cn$$−$$cTn2=(2cn$$−$$c)2=4c2n2+c2$$−$$4c2n$$ Sum $$=$$Σ$$Tn2=4c2+n(n+1)(2n+1)6+nc2$$−$$2c2n(n+1)=2c2n(n+1)(2n+1)+3nc2$$−$$6c2n(n+1)3=nc24n2+6n+2+3$$−$$6n$$−$$63=nc24n2$$−$$13$$

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If the sum of first n terms of an AP is cn², then the sum of squares of these n terms is $\frac{n (4 n^{2} - 1) c^{2}}{k}$ then k is

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Solution

$\begin{array}{l} S_{n} = {cn}^{2} \\ S_{n - 1} = c (n - 1)^{2} = {cn}^{2} + c - 2 cn \\ T_{n} = 2 cn - c \\ T_{n}^{2} = (2 cn - c)^{2} = 4 c^{2} n^{2} + c^{2} - 4 c^{2} n \\ Sum = {ΣT}_{n}^{2} \\ = \frac{4 c^{2} + n (n + 1) (2 n + 1)}{6} + {nc}^{2} - 2 c^{2} n (n + 1) \\ = \frac{2 c^{2} n (n + 1) (2 n + 1) + 3 {nc}^{2} - 6 c^{2} n (n + 1)}{3} \\ = \frac{{nc}^{2} (4 n^{2} + 6 n + 2 + 3 - 6 n - 6)}{3} = \frac{{nc}^{2} (4 n^{2} - 1)}{3} \end{array}$

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Special Series in Sequences and Series

Remember concepts with our Masterclasses.

If the sum of first n terms of an AP is cn2, then the sum of squares of these n terms is n4n2−1c2k then k is

Sn=cn2Sn−1=c(n−1)2=cn2+c−2cnTn=2cn−cTn2=(2cn−c)2=4c2n2+c2−4c2n Sum =ΣTn2=4c2+n(n+1)(2n+1)6+nc2−2c2n(n+1)=2c2n(n+1)(2n+1)+3nc2−6c2n(n+1)3=nc24n2+6n+2+3−6n−63=nc24n2−13

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If the sum of first n terms of an AP is cn², then the sum of squares of these n terms is $\frac{n (4 n^{2} - 1) c^{2}}{k}$ then k is