# Adding Multiples of Consecutive Odd Numbers by Mathematical Induction

(a)   Factorise $4x^3 + 18x^2 + 23x + 9$.

\begin{align} \displaystyle 4x^3 + 18x^2 + 23x + 9 &= 4x^3 + 4x^2 + 14x^2 + 23x + 9 \\ &= 4x^2 (x+1) + 14x^2 + 14x + 9x + 9 \\ &= 4x^2 (x+1) + 14x(x+1) + 9(x+1) \\ &= (x+1)(4x^2 + 14x + 9) \\ \end{align}

(b)   Hence, prove by mathematical induction that , for $x \ge 1$,
$1 \times 3 + 3 \times 5 + 5 \times 7 + \cdots + (2x-1)(2x+1) = \dfrac{x}{3}(4x^2 + 6x – 1)$.

Step 1: Prove the statement is true for $x=1$.
\begin{align} \displaystyle \text{LHS } &= 1 \times 3 = 3 \\ \text{RHS } &= \dfrac{1}{3} \times (4 \times 1^2 + 6 \times 1 – 1) = 3 \end{align}
Therefore it is true for $x=1$.
Step 2: Assume the statement is true for $x=k$.
That is, $1 \times 3 + 3 \times 5 + 5 \times 7 + \cdots + (2k-1)(2k+1) = \dfrac{k}{3}(4k^2 + 6k – 1)$
Step 3: Show the statement is true for $x = k+1$.
That is, $1 \times 3 + 3 \times 5 + 5 \times 7 + \cdots + (2k-1)(2k+1) + (2k+1)(2k+3) = \dfrac{k+1}{3}(4(k+1)^2 + 6(k+1) – 1)$
\begin{align} \displaystyle \require{color} \text{LHS } &= 1 \times 3 + 3 \times 5 + 5 \times 7 + \cdots + (2k-1)(2k+1) + (2k+1)(2k+3) \\ &= \dfrac{k}{3}(4k^2 + 6k – 1) + (2k+1)(2k+3) &\color{red} \text{by the assumption} \\ &= \dfrac{1}{3}(4k^3 + 6k^2 – k) + (4k^2 + 8k + 3) \\ &= \dfrac{1}{3}(4k^3 + 6k^2 – k + 12k^2 + 24k + 9) \\ &= \dfrac{1}{3}(4k^3 + 18k^2 + 23k + 9) \\ &= \dfrac{1}{3}(k+1)(4k^2 + 14k + 9) &\color{red} \text{by part (a)}\\ &= \dfrac{k+1}{3}(4k^2 + 14k + 9) \\ &= \dfrac{k+1}{3}(4k^2 + 8k + 4 + 6k + 6 -1) \\ &= \dfrac{k+1}{3}(4(k+1)^2 + 6(k+1) – 1) \\ &= \text{RHS} \\ \end{align}
So, the statement is true for $x = k$ it is also true for $x = k+1$.
It is true for $x =1$, so the statement is true for $x =2$ and so for $x = 3$ and so on.
Thus the statement is true for all $x \ge 1$ by the process of mathematical induction.