If S_1, S_2, S_3, dots, S_m are the sums of n | vedclass

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(A) The sum of $n$ terms of an $A.P.$ is given by $S = \frac{n}{2}[2a + (n - 1)d]$.For the $k$-th $A.P.$,the first term $a_k = k$ and the common diffe

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$\frac{1}{2}mn(mn + 1)$

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(A) The sum of $n$ terms of an $A.P.$ is given by $S = \frac{n}{2}[2a + (n - 1)d]$.For the $k$-th $A.P.$,the first term $a_k = k$ and the common difference $d_k = 2k - 1$.Thus,$S_k = \frac{n}{2}[2k + (n - 1)(2k - 1)] = \frac{n}{2}[2k + 2kn - n - 2k + 1] = \frac{n}{2}[2kn - n + 1]$.Now,we need to find the sum $\sum_{k=1}^{m} S_k = \sum_{k=1}^{m} \frac{n}{2}[2kn - n + 1]$.$= \frac{n}{2} [2n \sum_{k=1}^{m} k + \sum_{k=1}^{m} (1 - n)]$.$= \frac{n}{2} [2n \frac{m(m+1)}{2} + m(1 - n)]$.$= \frac{n}{2} [nm(m+1) + m - mn] = \frac{n}{2} [nm^2 + nm + m - mn] = \frac{n}{2} [nm^2 + m] = \frac{mn(mn + 1)}{2}$.

If $S_1, S_2, S_3, \dots, S_m$ are the sums of $n$ terms of $m$ arithmetic progressions $(A.P.)$ whose first terms are $1, 2, 3, \dots, m$ and common differences are $1, 3, 5, \dots, 2m - 1$ respectively,then $S_1 + S_2 + S_3 + \dots + S_m = $

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