Prove by mathematical induction that \(\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right)\), where \(n \in \mathbb{Z},n \geqslant 3\).

\(\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right)\)

show true for \(n = 3\)     (M1)

\({\text{LHS}} = \left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) = 1\) \(\,\,\,\) \({\text{RHS}} = \left( {\begin{array}{*{20}{c}} 3 \\ 3 \end{array}} \right) = 1\)     A1

hence true for \(n = 3\)

assume true for \(n = k:\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {k - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} k \\ 3 \end{array}} \right)\)     M1

consider for \(n = k + 1:\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {k - 1} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right)\)     (M1)

\( = \left( {\begin{array}{*{20}{c}} k \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right)\)     A1

\( = \frac{{k!}}{{(k - 3)!3!}} + \frac{{k!}}{{(k - 2)!2!}}\,\,\,\left( { = \frac{{k!}}{{3!}}\left[ {\frac{1}{{(k - 3)!}} + \frac{3}{{(k - 2)!}}} \right]} \right)\) or any correct expression with a visible common factor     (A1)

\( = \frac{{k!}}{{3!}}\left[ {\frac{{k - 2 + 3}}{{(k - 2)!}}} \right]\) or any correct expression with a common denominator     (A1)

\( = \frac{{k!}}{{3!}}\left[ {\frac{{k + 1}}{{(k - 2)!}}} \right]\)

Note:     At least one of the above three lines or equivalent must be seen.

\( = \frac{{(k + 1)!}}{{3!(k - 2)!}}\) or equivalent     A1

\( = \left( {\begin{array}{*{20}{c}} {k + 1} \\ 3 \end{array}} \right)\)

Result is true for \(k = 3\). If result is true for \(k\) it is true for \(k + 1\). Hence result is true for all \(k \geqslant 3\). Hence proved by induction.     R1

Note:     In order to award the R1 at least [5 marks] must have been awarded.

[9 marks]