Find the sum to n terms of the series 1 cdot | vedclass

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(A) The $r^{th}$ term of the series is $T_r = (2r-1)(2r+1)(2r+3)$.$T_r = (4r^2 - 1)(2r+3) = 8r^3 + 12r^2 - 2r - 3$.To find the sum $S_n = \sum_{r=1}^{

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$n(2n^3 + 8n^2 + 7n - 2)$

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(A) The $r^{th}$ term of the series is $T_r = (2r-1)(2r+1)(2r+3)$.$T_r = (4r^2 - 1)(2r+3) = 8r^3 + 12r^2 - 2r - 3$.To find the sum $S_n = \sum_{r=1}^{n} T_r$,we use the standard summation formulas:$S_n = 8 \sum r^3 + 12 \sum r^2 - 2 \sum r - \sum 3$.$S_n = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n$.$S_n = 2n^2(n+1)^2 + 2n(n+1)(2n+1) - n(n+1) - 3n$.$S_n = 2n^2(n^2+2n+1) + 2n(2n^2+3n+1) - n^2 - n - 3n$.$S_n = 2n^4 + 4n^3 + 2n^2 + 4n^3 + 6n^2 + 2n - n^2 - 4n$.$S_n = 2n^4 + 8n^3 + 7n^2 - 2n = n(2n^3 + 8n^2 + 7n - 2)$.

Find the sum to $n$ terms of the series $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$

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