4) $S_{4000}=\frac{4000(4000)}{2}=8\,002\,000$

Expert intro: This is an arithmetic-series subtraction problem. First find the total sum from 1 to 4000, then remove the multiples of 5.

Detailed walkthrough

We are given:

$S_{4000}=\frac{4000(4000)}{2}=8\,002\,000$

This is the sum of the numbers from 1 to 4000.

Now find the multiples of 5:

$5,10,15,\cdots,4000$

There are $800$ terms because

$4000=5n \Rightarrow n=800$

The sum of these multiples is

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

Now subtract:

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

Final answer

$6\,400\,000$

💡 Pitfall guide

Do not use $4000$ as the number of multiples of 5. The count is $800$, not $4000$. Also, be careful to include the first term 5 and the last term 4000 in the arithmetic sum.

🔄 Real-world variant

If the upper bound changes, the structure stays the same: total sum minus the sum of the chosen multiples. For multiples of another number $k$, replace $5$ by $k$ and count terms using $N=kn$.

🔍 Related terms

arithmetic progression, series sum, multiple