If the sum of the first n terms of an A.P. is | 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 $S_n - S_{n-1}$.$t_n = [cn^2 - cn] - [c(n-1)^

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

If the sum of the 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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