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(C) The $n^{th}$ term of the series is given by $t_n = \frac{1^3 + 2^3 + \dots + n^3}{1 + 3 + 5 + \dots + (2n-1)}$.Using the formulas for the sum of c

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$446$

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(C) The $n^{th}$ term of the series is given by $t_n = \frac{1^3 + 2^3 + \dots + n^3}{1 + 3 + 5 + \dots + (2n-1)}$.Using the formulas for the sum of cubes and the sum of the first $n$ odd numbers:$t_n = \frac{[\frac{n(n+1)}{2}]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4} = \frac{n^2}{4} + \frac{n}{2} + \frac{1}{4}$.Now,the sum of the first $16$ terms is $S_{16} = \sum_{n=1}^{16} t_n = \frac{1}{4} \sum_{n=1}^{16} n^2 + \frac{1}{2} \sum_{n=1}^{16} n + \frac{1}{4} \sum_{n=1}^{16} 1$.$S_{16} = \frac{1}{4} \left[ \frac{16(17)(33)}{6} \right] + \frac{1}{2} \left[ \frac{16(17)}{2} \right] + \frac{1}{4} (16)$.$S_{16} = \frac{1496}{4} + \frac{272}{4} + 4 = 374 + 68 + 4 = 446$.

What is the sum of the first $16$ terms of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$?

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