$\left[\begin{array}{l}\text { Arithmetic progression }\end{array}\right]$
[ Examples and counterexamples. Constructions $]$

Can the sum of 1000 consecutive odd numbers be the seventh power of a natural number?
Output the formula for the sum of a thousand consecutive odd numbers.

## Solution

Let (n-999), (n-997), ..., (n-1), (n+1), ..., (n+999) be a thousand consecutive odd numbers. Then their sum S= $(n-999)+(n-997)+\ldots+(n-1)+(n+1)+\ldots+(n+999)=1000$ n. If $n=10000$, then $S=1000 n=10000000=10^{7}$, that is, the sum $S$ is equal to the seventh power of a natural number.

## Answer

Yes.