(1+3+5+⋯+3983)/1992

The sum of the first n odd numbers = n^2

And the number of odd terms is given by [1 + n] / 2

So we have

[1 + 3983] / 2 = 1992 terms

So 

1992^2 / 1992 =

1992

(1+3+5+⋯+3983)/1992

$$\small{\text{
$
\dfrac{ 1+3+5+\cdots +3983} {1992} = \ ?
$
}}\\ \\
\small{\text{
The arithmetic Series $ 1+3+5+\cdots +3983 $ has $a_1 = 1$ and $a_n=3983$ and $d = 2$
}}\\
\small{\text{
The formula is $a_n = a_1+(n-1)\cdot d$ or
$ 3983 = 1 + (n-1)\cdot 2 $
}}\\
\small{\text{
So $n = 1992 $
}}\\
\small{\text{
The sum $ = 1+3+5+\cdots +3983 = n \left( \dfrac{a_1+a_n}{2} \right) = 1992\cdot \left( \dfrac{1+3983}{2} \right)
= 1992\cdot 1992
$
}}\\\\
\small{\text{
$
\dfrac{ 1+3+5+\cdots +3983} {1992} = \dfrac{1992\cdot 1992} {1992} = 1992$
}}\\ \\$$