If the sum of first n terms of an AP is cn2, then the sum of squares of these n terms is (2009) (a) 6n(4n2−1)c2 (b) 3n(4n2+1)c2(c) 3n(4n2−1)c2 (d) 6n(4n2+1)c2

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If the sum of first {n} terms of an {A}{P} is {c}{n}^{{2}}, th

Question

If the sum of first $n$ terms of an $A P$ is $c n^{2},$ then the sum of squares of these $n$ terms is (2009) (a) $\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{6}$ (b) $\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{3}$ (c) $\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{3}$ (d) $\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{6}$

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Charlotte Discussed

If the sum of first

n

terms of an

A P

c n^{2},

then the sum of squares of these

n

terms is (2009) (a)

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

(b)

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

(c)

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

(d)

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

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Question Text

If the sum of first

n

terms of an

A P

c n^{2},

then the sum of squares of these

n

terms is (2009) (a)

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

(b)

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

(c)

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

(d)

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