The last digit of $1+2+\cdots+n$ and a related squence

You may find the squence $\{a_{n}\}$ is periodic.

In fact, $\frac{(n+20)(n+20+1)}{2} = \frac{n^{2}+41n+420}{2} =\frac{n(n+1)}{2} +20n+210$ , which implies the last digit of $\frac{(n+20)(n+21)}{2}$ and $\frac{n(n+1)}{2}$ are equal.

So $a_{n+20}=a_{n}$ , and $S_{2018}=S_{18}+100S_{20}$ .

$S_{20}=\displaystyle\sum_{k=1}^{20}{a_{k}}=1+3+6+0+5+1+8+6+5+5+6+8+1+5+0+6+3+1+0+0=70$ and $S_{18}=70$ .

Thus $S_{2018}=\boxed{7070}$ .

Let $f(n) = \dfrac{n(n+1)}{2}$ ,

Now computing the first few values of $f(n)$ ,

$f(1) = 1 \implies a_0 = 1$

$f(2) = 3 \implies a_1 = 3$

$f(3) = 6 \implies a_2 = 6$

$f(4) = 10 \implies a_3 = 0$

You may notice that $a_1 - a_0 = 2, a_2 - a_1 = 3$ and so on.

This means that:

$a_{n + 1} = \left(a_n + (n + 1)\right)\pmod {10}$

$a_{n + 2} = \left(\left( a_{n + 1}\pmod{10} \right)+ (n + 2)\right)\pmod {10}$

Now generalizing:

$a_{n + k} = \left(\left(\left(\left(\left(a_n\pmod{10}\right) + (n + 1)\right)\pmod{10} + (n + 2)\right)\pmod{10} + \cdots\right) \pmod{10} + (n + k)\right)\pmod{10}$ .

Because we start with $a_0$ and go up to $a_{2018}$ we can now run the following program:

1
2
3
4
5
6
7
8

a = 1
sum = 1

for n in range(1, 2018):
    a = (a + (n + 1)) % 10
    sum  = sum + a

print(sum)

The final output of the variable sum , is $\boxed{7070}$ .