### Mathematical Induction

#### Understand the principle of mathematical induction

• (a) Prove, by mathematical induction, that for all positive integers n, $$1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + (2n - 1)(2n) = \frac{1}{3}n(n + 1)(4n - 1)$$.

(b) Hence, according to the formula $$1 + 2 + 3 + \cdots + n = \frac{1}{2}n(n + 1)$$, deduce the formula for evaluating $${2^2} + {4^2} + {6^2} + \cdots + {(2n)^2}$$.
• (a) Let P(n) be ‘$$1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + (2n - 1)(2n) = \frac{1}{3}n(n + 1)(4n - 1)$$’.
When $$n = 1$$, L.H.S.$$= 1 \times 2 = 2$$

R.H.S.$$= \frac{1}{3} \times 1 \times 2 \times 3 = 2$$

∴P(1) is true.

Assume that P(k) is true, where k is a positive integer.

i.e. $$1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + (2k - 1)(2k) = \frac{1}{3}k(k + 1)(4k - 1)$$

then $$\begin{array}{l}1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + (2k - 1)(2k) + [2(k + 1) - 1][2(k + 1)]\\ = \frac{1}{3}k(k + 1)(4k - 1) + 2(k + 1)(2k + 1)\\ = \frac{1}{3}(k + 1)[k(4k - 1) + 6(2k + 1)]\\ = \frac{1}{3}(k + 1)(4{k^2} + 11k + 6)\\ = \frac{1}{3}(k + 1)(k + 2)(4k + 3)\\ = \frac{1}{3}(k + 1)[(k + 1) + 1][4(k + 1) - 1]\end{array}$$

∴ $$P(k + 1)$$ is true.

According to mathematical induction, P(n) is true for all positive integers n.

(b)

$$\begin{array}{1}{2^2} + {4^2} + {6^2} + \cdots + {(2n)^2} = (1 + 1) \times 2 + (1 + 3) \times 4 + (1 + 5) \times 6 + \cdots + [1 + (2n - 1)](2n)\\ = 2 + 1 \times 2 + 4 + 3 \times 4 + 6 + 5 \times 6 + \cdots + 2n + (2n - 1)(2n)\\ = [1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + (2n - 1)(2n)] + (2 + 4 + 6 + \cdots + 2n)\\ = \frac{1}{3}n(n + 1)(4n - 1) + 2(1 + 2 + 3 + \cdots + n)\\ = \frac{1}{3}n(n + 1)(4n - 1) + 2[\frac{1}{2}n(n + 1)]\\ = \frac{1}{3}n(n + 1)[(4n - 1) + 3]\\ = \frac{1}{3}n(n + 1)(4n + 2)\\ = \underline{\underline {\frac{2}{3}n(n + 1)(2n + 1)}} \end{array}$$