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

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\left(\begin{array}{c}2\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ 2\end{array}\right)+\dots +\left(\begin{array}{c}n-1\\ 2\end{array}\right)=\left(\begin{array}{c}n\\ 3\end{array}\right)$

show true for $n=3$     (M1)

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

hence true for $n=3$

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

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

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

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

Result is true for $k=3$. If result is true for $k$ it is true for $k+1$. Hence result is true for all $k⩾3$. Hence proved by induction.     R1

 

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

 

[9 marks]

[N/A]