If the sum of first n terms of an A.P. is cn( | vedclass

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(B) Given the sum of the first $n$ terms $S_n = cn(n-1) = cn^2 - cn$.The $n^{th}$ term $t_n$ is given by $S_n - S_{n-1}$.$S_{n-1} = c(n-1)(n-2) = c(n^

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

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(B) Given the sum of the first $n$ terms $S_n = cn(n-1) = cn^2 - cn$.The $n^{th}$ term $t_n$ is given by $S_n - S_{n-1}$.$S_{n-1} = c(n-1)(n-2) = c(n^2 - 3n + 2)$.$t_n = (cn^2 - cn) - (cn^2 - 3cn + 2c) = 2cn - 2c = 2c(n-1)$.We need to find the sum of the squares of these terms,i.e.,$\sum_{k=1}^{n} (t_k)^2$.$t_k = 2c(k-1)$.$(t_k)^2 = 4c^2(k-1)^2 = 4c^2(k^2 - 2k + 1)$.Sum $= \sum_{k=1}^{n} 4c^2(k^2 - 2k + 1) = 4c^2 [\sum k^2 - 2\sum k + \sum 1]$.Using standard summation formulas:Sum $= 4c^2 [\frac{n(n+1)(2n+1)}{6} - 2\frac{n(n+1)}{2} + n]$.Sum $= 4c^2 [\frac{n(n+1)(2n+1) - 6n(n+1) + 6n}{6}]$.Sum $= \frac{4c^2}{6} [n(2n^2 + 3n + 1 - 6n - 6 + 6)] = \frac{2c^2}{3} [n(2n^2 - 3n + 1)]$.Sum $= \frac{2c^2}{3} [n(n-1)(2n-1)] = \frac{2}{3}c^2n(n-1)(2n-1)$.

If the sum of first $n$ terms of an $A.P.$ is $cn(n - 1)$,where $c \neq 0$,then the sum of the squares of these terms is

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