Answer
$a_{n}=n^{2}-3$
Work Step by Step
second term = first +3
third term = second +5
fourth term = third +7
fifth term = fourth +9
The nth odd number added can be written as $2n-1$
$a_{1}=-2$
$a_{2}=a_{1}+(2\cdot 2-1)$
$a_{3}=a_{2}+(3\cdot 2-1)$
$=a_{1}+(2\cdot 2-1)+(3\cdot 2-1)$
$a_{4}=a_{3}+(4\cdot 2-1)$
$=a_{1}+(2\cdot 2-1)+(3\cdot 2-1)+(4\cdot 2-1)$
$a_{5}=a_{4}+(5\cdot 2-1)$
$=a_{1}+(2\cdot 2-1)+(3\cdot 2-1)+(4\cdot 2-1)+(5\cdot 2-1)$
$...$
$a_{n}=a_{1}+(2\cdot 2-1)+(3\cdot 2-1)+(4\cdot 2-1)+(5\cdot 2-1)+...(n\cdot 2-1)$
... there are n-1 terms of -1,
$=a_{1}+2(2+3+4+..+n)-(n-1)$
... The parentheses contain the sum of integers from 1 to n, minus 1
$(2+3+4+..+n)=\displaystyle \frac{n(n+1)}{2}-1$
$a_{n}=a_{1}+2(\displaystyle \frac{n(n+1)}{2} -1)-(n-1)$
$a_{n}=-2+n^{2}+n-2-n+1$
$a_{n}=n^{2}-3$
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Alternatively, you may have recognized the pattern
$a_{1}=1^{2}-3$
$a_{2}=2^{2}-3$
$a_{3}=3^{2}-3$
$a_{4}=4^{2}-3 ...$
leading to the same answer
