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(A) The $n$-th term of the series is $T_n = (2n-1)(2n+1)(2n+3) = (4n^2-1)(2n+3) = 8n^3 + 12n^2 - 2n - 3$. The sum of $n$ terms is $S_n = \sum_{k=1}^{n

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$9$

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(A) The $n$-th term of the series is $T_n = (2n-1)(2n+1)(2n+3) = (4n^2-1)(2n+3) = 8n^3 + 12n^2 - 2n - 3$. The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 8 \sum k^3 + 12 \sum k^2 - 2 \sum k - 3 \sum 1$. $S_n = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n$. $S_n = 2n^2(n+1)^2 + 2n(n+1)(2n+1) - n(n+1) - 3n$. $S_n = n(n+1) [2n(n+1) + 2(2n+1) - 1] - 3n$. $S_n = n(n+1) [2n^2 + 2n + 4n + 2 - 1] - 3n = n(n+1)(2n^2 + 6n + 1) - 3n$. Comparing with $n(n+1)f(n) - 3n$,we get $f(n) = 2n^2 + 6n + 1$. Therefore,$f(1) = 2(1)^2 + 6(1) + 1 = 2 + 6 + 1 = 9$.

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

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