Prove $2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (n^2+1)n! = n(n+1)!$. Step 1 Show it is true for $n=1$. \begin{align} &\text{LHS} = (1^2+1) \times 1! = 2 \\ &\text{RHS} = 1 \times (1+1)! = 2 \\ &\text{LHS} = \text{RHS} \\ &\text{Therefore it is […] # Determining Initial Values | Principles of Mathematical Induction Prove \( 1+3+5+\cdots+(2n+1) = (n+1)^2. Step 1 Show it is true for $n=0$ by mathematical induction. \begin{align} &\text{LHS} = 2 \times 0 +1 = 1 \\ &\text{RHS} = (0+1)^2 = 1 \\ &\text{LHS} = \text{RHS} \\ &\text{Therefore, it is true for } n=0 \end{align} Step 2 Assume that it is […] Example 1 Factorise $(x-1)(x-3)(x+2)(x+4)+24$. \require{AMSsymbols} \begin{align} &= (x-1)(x+2) \times (x-3)(x+4) + 24 \\ &= (x^2+x-2) \times (x^2+x-12) +24 \\ &= \left[(x^2+x)-2\right] \times \left[(x^2+x)-12\right] +24 \\ &= (x^2+x)^2-14(x^2+x)+24+24 \\ &= (x^2+x)^2\bbox[aqua]{-14(x^2+x)}+48 \end{align} $\require{AMSsymbols} \begin{array} {ccr} &\bbox[yellow]{x^2+x} &\bbox[pink,3px]{-6} &\bbox[pink]{-6(x^2+x)} \\ &\bbox[pink]{x^2+x} &\bbox[yellow,3px]{-8} &\bbox[yellow]{-8(x^2+x)} \\ \hline &&&\bbox[aqua]{-14(x^2+x)} \end{array}$ \require{AMSsymbols} \begin{align} &= […] # Factorising Quadratics with Six Terms of \( x and $y$ such as $x^2 + 2xy + 5x + 5y + y^2 + 6$

Sometimes students may encounter complex quadratics factorise involving $x^2, y^2, xy, x$ and $y$. Consider factorising by only either $x$ or $y$. The following examples take through factorising $y$ first, then $x$. Example 1 Factorise \( x^2 + 2xy + 5x + y^2+ 5y […]