Double counting once

The sum of first $n$ natural numbers is given by $\dfrac{n(n + 1)}{2}$ . Assume that this is equal to $2018$ .

$\begin{aligned} \dfrac{n(n + 1)}{2} & = 2018 \\ n^2 + n & = 2018 \times 2 \\ n^2 + n & - 4036 = 0 \\ \implies n & = \dfrac{1 + \sqrt{16145}}{2} \qquad (\text{We can exclude }n = \dfrac{1 - \sqrt{16145}}{2}\text{ as n cannot be negative}) \\ \implies n & = 64.03 \\ \end{aligned}$

• Take $n = 64$ , we will get $\dfrac{n(n + 1)}{2} = 2080$ which is more than $2018$ so we should not take $n = 64$ .
• Take $n = 63$ , we will get $\dfrac{n(n + 1)}{2} = 2016$ which is less than $2018$ so we get $n = 63$ .

$\therefore$ The number which is added twice is $= 2018 - 2016 = 2$

Let the double counted number be $k$ .

$1+2+3+...n+k$ = 2018)

$\frac {n(n+1)}{2} + k =2018$

(The first term of the LHS of the above equation is called a triangular number)

The only triangular number a little less than 2018 is 2016 ( $n=63$ ). Hence, the number $2$ was counted twice.