Find it

The expression, $\frac { 1+3+5+...+3983 }{ 1992 }$ , can be solved easily by applying knowledge of the arithmetic series.

First, the expression above the numerator, $1+3+5+...+3983$ , is an example of an arithmetic series. Recall that to solve for the arithmetic series, the formula, ${ S }_{ n }=\frac { n(a_{ n }+a_{ 1 }) }{ 2 }$ , can be applied; however, before we can apply the formula, we must find the common difference d and the number of terms n.

To find the common difference, subtract the second term from the first term. Generally, the common difference of any arithmetic series or sequence is $d=\frac { a_{ n } }{ a_{ n-1 } }$ . Applying the formula, we can get 2 as the common difference.

To find the number of terms, we can rewrite first the arithmetic series as an arithmetic sequence. Then, the formula for arithmetic sequence, $a_{ n }=a_{ 1 } + ( n-1 ) d$ , can be applied. Substituting the terms, we get: $3983=1+(n-1)(2)$ . Manipulating the expression, we get $n=1992$ .

Then, we can apply the formula for the arithmetic series.

${ S }_{ n }=\frac { 1992(3983+1) }{ 2 }$

${ S }_{ n }=\frac { 1992(3984) }{ 2 }$

${ S }_{ n }=1992(1992)$

Next, we can rewrite the expression, $\frac { 1+3+5+...+3983 }{ 1992 }$ , as $\frac { 1992(1992) }{ 1992 }$ . The two 1992 can be cancelled out.

Finally, we are left with 1992, which is the answer to the expression, $\frac { 1+3+5+...+3983 }{ 1992 }$ .

$\frac { 1+3+5+...+3983 }{ 1992 }=1992$