4) $S_{4000}=\frac{4000(4000)}{2}

Key concept: This problem combines two arithmetic-series sums: the sum of the first 4000 integers and the sum of the multiples of 5 up to 4000. Subtracting them gives the sum of the numbers that are not multiples of 5.

Step by step

We want the sum of all integers from 1 to 4000, then subtract the sum of the multiples of 5.

1) Sum of all numbers from 1 to 4000

Use the formula

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

So

$S_{4000}=\frac{4000(4000+1)}{2}=\frac{4000\cdot 4001}{2}=8\,002\,000$

2) Sum of the multiples of 5 up to 4000

The multiples of 5 are

$5,10,15,\ldots,4000$

This is an arithmetic sequence with:

• first term $a=5$
• last term $l=4000$
• number of terms $n$ found by $5n=4000$, so $n=800$

Now use

$S_n=\frac{n}{2}(a+l)$

So

$S_{800}=\frac{800}{2}(5+4000)=400\cdot 4005=1\,602\,000$

3) Subtract

$8\,002\,000-1\,602\,000=6\,400\,000$

Answer

$6\,400\,000$

Pitfall alert

Do not count the multiples of 5 as 4000 terms. There are only 800 of them, because $4000\div 5=800$. Also, remember to use the arithmetic-series formula with the correct first and last terms.

Try different conditions

If the upper limit changes from 4000 to another number $N$, the same method works: find $\frac{N(N+1)}{2}$ for all integers, then subtract the arithmetic-series sum of the multiples of 5 up to $N$.

Further reading

arithmetic series, sum formula, multiples of 5