Question 127
A technique widely used to compute sums is the Telescoping Sum. It consists of "decomposing" the terms of a sum into parts that cancel. 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 add consecutive odd numbers. Let us see:
a) Counting the odd numbers one by one, starting from 1, verify that the number in position $m$ is equal to $m^{2}-(m-1)^{2}$.
b) Compute the sum of all odd numbers between 1000 and 2014.
$$
\begin{aligned}
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}= \\
& \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)= \\
& \frac{1}{1}-\frac{1}{5}= \\
& \frac{4}{5}
\end{aligned}
$$
With this technique, we can find a way to add consecutive odd numbers. Let us see:
a) Counting the odd numbers one by one, starting from 1, verify that the number in position $m$ is equal to $m^{2}-(m-1)^{2}$.
b) Compute the sum of all odd numbers between 1000 and 2014.