91Ó°ÊÓ

First, we will write down the formula for the sum of squares for the first 1000 natural numbers as follows: $$ \sum_{m=1}^{1000} m^{2} $$

Now, we will define the given expression containing both even and odd powers of m as follows: $$ \sum_{m=0}^{999}\left(m^{2}+2 m+1\right) $$

Next, we will expand the given expression and break it down into individual summations as follows: $$ \sum_{m=0}^{999}(m^2)+\sum_{m=0}^{999}(2m)+\sum_{m=0}^{999}(1) $$

Observe that the first term in the expanded expression, \(\sum_{m=0}^{999}m^2\), is similar to the sum of squares formula but only goes up to 999 instead of 1000. This means it contains all the squared terms except for 1000². Now, let's look at the second term, \(\sum_{m=0}^{999}(2m)\), which is the sum of all even numbers from 0 to 1998. Notice that this sum includes all the even numbers between the squares with a difference of 2, which means it fills the gap between the squared terms. Finally, the third term, \(\sum_{m=0}^{999}(1)\), is simply the sum of 1000 ones which is equal to 1000.

To show that the sum of squares formula is equal to the given expression, we will demonstrate that the sum of the squares from 1 to 1000 can be rearranged to form the even and odd powered terms of the given expression: $$ (1^2 + 2^2 + \cdots + 1000^2) = (0^2 + 1^2 + \cdots + 999^2) + (0 + 2 + 4 + \cdots + 1998) + (1 + 1 + \cdots + 1) $$ Now, the sum of the first \(1000\) ones is equal to \(1000\). Thus, adding the sum of even numbers from 0 to 1998 to the sum of 1000 ones gives us the sum of odd numbers from 1 to 1999: $$ (0 + 2 + 4 + \cdots + 1998) + (1 + 1 + \cdots + 1) = (1 + 3 + 5 + \cdots + 1999) $$ Adding the sum of even and odd numbers between squares and putting this summed result in place gives: $$ (1^2 + 2^2 + \cdots + 1000^2) = (0^2 + 1^2 + \cdots + 999^2) + (1 + 3 + 5 + \cdots + 1999) $$ Thus, we have shown that the sum of squares for the first 1000 natural numbers can be represented as the given expression containing both even and odd powers of m: $$ \sum_{m=1}^{1000} m^{2}=\sum_{m=0}^{999}\left(m^{2}+2 m+1\right) $$