Question
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{\text{n}(\text{n}-2)}{2}$
$\text{a}_\text{n}=\frac{\text{n}(\text{n}-2)}{2}$
✓
Answer
$\text{a}_\text{n}=\frac{\text{n}(\text{n}-2)}{2}$
Let $n = 1, 2, 3, 4, 5,$ then
$\text{a}_1=\frac{1(1-2)}{2}=\frac{1\times(-1)}{2}=\frac{-1}{2}$
$\text{a}_2=\frac{1(2-2)}{2}=\frac{2\times0}{2}=0$
$\text{a}_3=\frac{1(3-2)}{2}=\frac{3\times1}{2}=\frac{3}{2}$
$\text{a}_4=\frac{4(4-2)}{2}=\frac{4\times2}{2}=4$
$\text{a}_5=\frac{5(5-2)}{2}=\frac{5\times3}{2}=\frac{15}{2}$
Let $n = 1, 2, 3, 4, 5,$ then
$\text{a}_1=\frac{1(1-2)}{2}=\frac{1\times(-1)}{2}=\frac{-1}{2}$
$\text{a}_2=\frac{1(2-2)}{2}=\frac{2\times0}{2}=0$
$\text{a}_3=\frac{1(3-2)}{2}=\frac{3\times1}{2}=\frac{3}{2}$
$\text{a}_4=\frac{4(4-2)}{2}=\frac{4\times2}{2}=4$
$\text{a}_5=\frac{5(5-2)}{2}=\frac{5\times3}{2}=\frac{15}{2}$
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