Question
Write the first five terms of the sequence whose nth term is $a _ { n } = \frac { 2 n - 3 } { 6 }$.
Putting n = 1, 2, 3, 4 and 5, we get,
$a _ { 1 } = \frac { 2 \times 1 - 3 } { 6 } = \frac { 2 - 3 } { 6 } = \frac { - 1 } { 6 }$
$a _ { 2 } = \frac { 2 \times 2 - 3 } { 6 } = \frac { 4 - 3 } { 6 } = \frac { 1 } { 6 }$
$a _ { 3 } = \frac { 2 \times 3 - 3 } { 6 } = \frac { 6 - 3 } { 6 } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
$a _ { 4 } = \frac { 2 \times 4 - 3 } { 6 } = \frac { 8 - 3 } { 6 } = \frac { 5 } { 6 }$
$a _ { 5 } = \frac { 2 \times 5 - 3 } { 6 } = \frac { 10 - 3 } { 6 } = \frac { 7 } { 6 }$
Therefore, the first five terms are $\frac { - 1 } { 6 } , \frac { 1 } { 6 } , \frac { 1 } { 2 } , \frac { 5 } { 6 }$ and $\frac { 7 } { 6 }.$
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$\text{x}^2\text{e}^\text{x}\log\text{x}$
$\frac12,\frac13,\frac29,\frac{4}{27}\dots$