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.

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