HSC Mathematics Advanced, Extension 1, and Extension 2/3-Unit/HSC/Induction

Induction is a form of proof useful for proving equations involving non-closed expressions (i.e., expressions with $$n$$ terms; sequences).

Explanation
Induction involves first proving that the equation is true for $$n = 1$$, then proving true for $$n = k + 1$$ (assuming for the purpose of the proof that the equation holds true for $$n = k$$). Since it is true for $$n = k$$ and true for $$n = k+1$$, and also true for $$n = 1$$, it is true for $$n = 2$$. It follows that it is true for all positive integers $$n$$.

Proving the formula for the sum of a series
Q: Prove by mathematical induction that for all integers $$n \ge 1$$,
 * $$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = (1+2+3+....n)^2$$

A:  When $$n = 1$$, $$1^3 = 1 = \tfrac{1}{4} (1)^2((1)+1)^2 = \tfrac{1}{4}(4) = 1$$, so it is true for $$n = 1$$ Suppose that the statement is true for $$k, k \in \mathbb{N}$$. That is, suppose that $$1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 = \tfrac{1}{4} k^2 (k+1)^2$$. This is sometimes called the induction hypothesis. Then prove the statement for $$n = k+1$$ (that is, prove that $$1^3 + 2^3 + 3^3 + \cdots + (k+1)^3 = \tfrac{1}{4} (k+1)^2(k+2)^2$$:

\begin{align} \mbox{LHS} & = 1^3 + 2^3 + 3^3 + 4+3 + \cdots + k^3 + (k+1)^3 \\ & = \tfrac{1}{4} k^2 (k+1)^2 + (k+1)^3               & \mbox{ (by the induction hypothesis)} \\ & = \tfrac{1}{4}(k+1)^2(k^2 + 4(k+1)) \\ & = \tfrac{1}{4}(k+1)^2(k^2 + 4k + 4) \\ & = \tfrac{1}{4}(k+1)^2(k+2)^2 \\ & = \mbox{RHS} \end{align} $$  It follows from parts 1 and 2 by mathematical induction that the statement is true for all positive integers $$n$$. 