1310.11 – Evens Minus Odds

The sum of the first 80 positive odd integers subtracted from the sum of the first 80 positive even integers is:

  1. 0

  2. 20

  3. 40

  4. 60

  5. 80

Solution

Here's a nice method:

E=2+4+…+160E = 2 + 4 + \ldots + 160

O=1+3+…+159O = 1 + 3 + \ldots + 159

We want:

E−O=(2−1)+(4−3)+…+(160−159)=1+1+…+1=80\begin{align*}E - O &= (2-1) + (4-3) + \ldots + (160-159) \\&= 1 + 1 + \ldots + 1 \\&= 80\end{align*}

Here's another method:

The sum of an arithmetic series is n2⋅(f−l)\dfrac{n}{2} \cdot (f - l) where there are nn terms, ff is the first, and ll is the last.

So E=802⋅(2+160)=6480E = \dfrac{80}{2} \cdot (2 + 160) = 6480, and O=802⋅(1+159)=6400O = \dfrac{80}{2} \cdot (1 + 159) = 6400.

Again, the difference is 80. So the answer is (e).