BRILLIAthon Day $2$ , Problem $2$ of $2$

Both even and odd number sequences are arithmetic sequences whose terms have the equation $a_n = a_1 + d(n - 1)$ and whose sums have the equation $S_n = \frac{1}{2}n(a_1 + a_n)$ for first term $a_1$ and difference $d$ .

For the first $2000$ even numbers, $a_1 = 2$ , $d = 2$ , and $a_{2000} = a_1 + d(2000 - 1) = 2 + 2(2000 - 1) = 4000$ , which means that its sum is $S_{\text{even}} = \frac{1}{2}n(a_1 + a_n) = \frac{1}{2} \cdot 2000 \cdot (2 + 4000) = 4002000$ .

For the first $2000$ odd numbers, $a_1 = 1$ , $d = 2$ , and $a_{2000} = a_1 + d(2000 - 1) = 1 + 2(2000 - 1) = 3999$ , which means that its sum is $S_{\text{odd}} = \frac{1}{2}n(a_1 + a_n) = \frac{1}{2} \cdot 2000 \cdot (1 + 3999) = 4000000$ .

Therefore, $S_{\text{even}} - S_{\text{odd}} = 4002000 - 4000000 = \boxed{2000}$ .

$\begin{aligned} S_{\text{even}} & = 2 + 4 + 6 + \cdots + 4000 \\ S_{\text{odd}} & = 1 + 3 + 5 + \cdots + 3999 \\ S_{\text{even}} - S_{\text{odd}} & = \underbrace{1 + 1 + 1 + \cdots 1}_{\text{Number of 1s = }2000} = 2000 \end{aligned}$

Therefore the answer is $\boxed{2000}$ .

Important Points $\Downarrow$

$\Rightarrow$ 0 is an even number

$\Rightarrow$ There are two parts of the problem, so we should solve this step-by-step

Find the sum of the first $2000$ even numbers subtracted from the sum of the first $2000$ odd numbers.

• This can be rewritten as $\Rightarrow$ First $2000$ Odd Numbers $-$ First $2000$ Even Numbers

• Let's use our logic here $\Downarrow$

• We know that the first even number, $0$ is $1$ less than the first odd number, $1$

• We notice that the second even number, $2$ is also $1$ less than the second odd number, $3$

• This rule applies to all numbers, as an odd number is an even number $+1$

$\begin{aligned} x &= First \ 2000 \ Odd \ Numbers \ - \ First \ 2000 \ Even \ Numbers \\ &= First \ 2000 \ Odd \ Numbers \ - \ (First \ 2000 \ Odd \ Numbers \ - \ 2000) \\ &= 2000 \end{aligned}$