A technique widely used to calculate summations is the Telescoping Sum. It consists of "decomposing" the terms of a sum into parts that cancel each other out. For example,

$$
\begin{aligned}
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}= \\
& \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)= \\
& \frac{1}{1}-\frac{1}{5}= \\
& \frac{4}{5}
\end{aligned}
$$

With this technique, we can find a way to sum consecutive odd numbers. Let's see:

a) Counting the odd numbers one by one starting from 1, verify that the number at position $m$ is equal to $m^{2}-(m-1)^{2}$.

b) Calculate the sum of all odd numbers between 1000 and 2014.
Solution

a) See that the first odd number is $2 \cdot 1-1$ and, knowing that odd numbers increase by 2, we can conclude that the odd number at position $m$ in our count is

$$
\begin{aligned}
2 \cdot 1-1+\underbrace{2+2+\ldots+2}_{m-1 \text { times }} & =2 \cdot 1-1+2(m-1) \\
& =2 m-1
\end{aligned}
$$

To verify that it matches the number in item a), we just need to calculate

$$
m^{2}-(m-1)^{2}=m^{2}-\left(m^{2}-2 m-1\right)=2 m-1
$$

b) We want to sum the odd numbers from $1001=2 \cdot 501-1$ to $2013=2 \cdot 1007-1$. Using the expression from item a), we have

$$
\begin{aligned}
1001 & =501^{2}-500^{2} \\
1003 & =502^{2}-501^{2} \\
1005 & =503^{2}-502^{2} \\
& \cdots \\
2011 & =1006^{2}-1005^{2} \\
2013 & =1007^{2}-1006^{2}
\end{aligned}
$$

Adding everything, we see that all numbers from $501^{2}$ to $1006^{2}$ are canceled. Thus, the result is:

$$
\begin{aligned}
1001+1003+\ldots+2013 & =1007^{2}-500^{2} \\
& =(1007-500)(1007+500) \\
& =507 \cdot 1507 \\
& =764049
\end{aligned}
$$

So the sum of the odd numbers between 1000 and 2014 is 763048.