The arithmetic progression ${2}$, ${4}$, ${6}$, ${8}$, ${10}$, ${12}$, ${...}$ is arra

Question

The arithmetic progression ${2}$, ${4}$, ${6}$, ${8}$, ${10}$, ${12}$, ${...}$ is arranged in rows in the following way: 1st Row: $ \qquad\qquad\; {2}$ 2nd Row: $ \qquad\quad\;\;\; {4}$, ${6}$ 3rd Row: $ \qquad\qquad {8}$, ${10}$, ${12}$ 4th Row: $ \qquad\qquad {14}$, ${16}$, ${18}$, ${20}$ ${...}$ Show that the first term in the $n^{\mathrm{th}}$ row is ${{n}^{2}}-n+2$.[2] Find the sum of all the terms from the ${{1}^{\mathrm{st}}}$ row to the $\left( n-1 \right)^{\mathrm{th}}$ row.[3]

Accepted Answer

(i) First term $=n^2-n+2$ (ii) $\frac{n(n-1)(n^2-n+2)}{4}$

2009 DHS P1 Q9

2009 DHS P1 Q9

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