What is the arithmetic mean of the first n te | vedclass

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(D) The $n$-th term of the sequence is given by $T_n = (2n - 1)(2n + 1)(2n + 3)$.Expanding this,we get $T_n = (4n^2 - 1)(2n + 3) = 8n^3 + 12n^2 - 2n -

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

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(D) The $n$-th term of the sequence is given by $T_n = (2n - 1)(2n + 1)(2n + 3)$.Expanding this,we get $T_n = (4n^2 - 1)(2n + 3) = 8n^3 + 12n^2 - 2n - 3$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 8\sum k^3 + 12\sum k^2 - 2\sum k - \sum 3$.Using the standard summation formulas:$S_n = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n$.$S_n = 2n^2(n^2 + 2n + 1) + 2n(2n^2 + 3n + 1) - n(n+1) - 3n$.$S_n = 2n^4 + 4n^3 + 2n^2 + 4n^3 + 6n^2 + 2n - n^2 - n - 3n$.$S_n = 2n^4 + 8n^3 + 7n^2 - 2n$.The arithmetic mean is $\frac{S_n}{n} = \frac{2n^4 + 8n^3 + 7n^2 - 2n}{n} = 2n^3 + 8n^2 + 7n - 2$.

What is the arithmetic mean of the first $n$ terms of the sequence $1 \times 3 \times 5, 3 \times 5 \times 7, 5 \times 7 \times 9, \dots$?

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