The problem asks to find the sum $S_n$ of the following two series: (1) $1 \cdot 3, 3 \cdot 5, 5 \cdot 7, \dots, (2n-1)(2n+1)$ (2) $1^2 \cdot 2, 2^2 \cdot 3, 3^2 \cdot 4, \dots, n^2(n+1)$

1. Problem Description

The problem asks to find the sum $S_n$ of the following two series:

(1) $1 \cdot 3, 3 \cdot 5, 5 \cdot 7, \dots, (2n-1)(2n+1)$

(2) $1^2 \cdot 2, 2^2 \cdot 3, 3^2 \cdot 4, \dots, n^2(n+1)$

2. Solution Steps

(1) Let $a_k = (2k-1)(2k+1)$ be the $k$-th term of the series. Then

$a_k = 4k^2 - 1$

We want to find the sum $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (4k^2 - 1)$.

Using the formulas:

$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^n 1 = n$

Then, $S_n = \sum_{k=1}^n (4k^2 - 1) = 4 \sum_{k=1}^n k^2 - \sum_{k=1}^n 1 = 4 \cdot \frac{n(n+1)(2n+1)}{6} - n = \frac{2n(n+1)(2n+1)}{3} - n$.

$S_n = \frac{2n(n+1)(2n+1) - 3n}{3} = \frac{n(2(n+1)(2n+1) - 3)}{3} = \frac{n(2(2n^2+3n+1) - 3)}{3} = \frac{n(4n^2 + 6n + 2 - 3)}{3} = \frac{n(4n^2 + 6n - 1)}{3}$.

Therefore, $S_n = \frac{4n^3 + 6n^2 - n}{3}$.

(2) Let $a_k = k^2(k+1)$ be the $k$-th term of the series. Then $a_k = k^3 + k^2$.

We want to find the sum $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^3 + k^2)$.

Using the formulas:

$\sum_{k=1}^n k^3 = (\frac{n(n+1)}{2})^2$

$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Then, $S_n = \sum_{k=1}^n (k^3 + k^2) = \sum_{k=1}^n k^3 + \sum_{k=1}^n k^2 = (\frac{n(n+1)}{2})^2 + \frac{n(n+1)(2n+1)}{6} = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}$.

$S_n = \frac{3n^2(n+1)^2 + 2n(n+1)(2n+1)}{12} = \frac{n(n+1)(3n(n+1) + 2(2n+1))}{12} = \frac{n(n+1)(3n^2 + 3n + 4n + 2)}{12} = \frac{n(n+1)(3n^2 + 7n + 2)}{12}$.

$S_n = \frac{n(n+1)(3n+1)(n+2)}{12}$.

$S_n = \frac{n(n+1)(3n^2+7n+2)}{12} = \frac{3n^4 + 10n^3 + 9n^2 + 2n}{12}$.

3. Final Answer

(1) $S_n = \frac{4n^3 + 6n^2 - n}{3}$

(2) $S_n = \frac{3n^4 + 10n^3 + 9n^2 + 2n}{12} = \frac{n(n+1)(3n+1)(n+2)}{12}$