Question

For every positive integer n, the nth term of a sequence is given by a_{n} = \frac{1}{n} - \frac{1}{n + 1}. What is the sum of the first 100 terms of this sequence?

Answer 1

The sum telescopes nicely:

$\displaystyle\sum_{n=1}^{100}a_n=\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{99}-\frac1{100}\right)+\left(\frac1{100}-\frac1{101}\right)$

$\implies\displaystyle\sum_{n=1}^{100}a_n=1-\frac1{101}=\boxed{\frac{100}{101}}$