Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$
For $n = 1$,we have
$P(1): 1 \cdot 3 = 3 = \frac{1(4(1)^2 + 6(1) - 1)}{3} = \frac{4 + 6 - 1}{3} = \frac{9}{3} = 3$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2k - 1)(2k + 1) = \frac{k(4k^2 + 6k - 1)}{3}$ $(i)$
We shall now prove that $P(k + 1)$ is true.
Consider the sum up to $(k + 1)$ terms:
$(1 \cdot 3 + 3 \cdot 5 + \ldots + (2k - 1)(2k + 1)) + (2(k + 1) - 1)(2(k + 1) + 1)$
$= \frac{k(4k^2 + 6k - 1)}{3} + (2k + 1)(2k + 3)$ [Using $(i)$]
$= \frac{4k^3 + 6k^2 - k + 3(4k^2 + 8k + 3)}{3}$
$= \frac{4k^3 + 6k^2 - k + 12k^2 + 24k + 9}{3}$
$= \frac{4k^3 + 18k^2 + 23k + 9}{3}$
$= \frac{(k + 1)(4(k + 1)^2 + 6(k + 1) - 1)}{3}$
Thus,$P(k + 1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.