Hidden Telescoping Sum

Here’s an exercise I proposed in an online competition.

You have the following sequence:

a_1 = \frac{1}{70}, a_{n+1} = a_n^2 + a_n

Find the integer part of:

\sum_{i=1}^{100} \frac{1}{a_i + 1}

Now, notice that

\frac{1}{a_{n+1}} = \frac{1}{a_n(a_n + 1)} = \frac{1}{a_n} - \frac{1}{a_n+1}


\frac{1}{a_n} - \frac{1}{a_{n+1}} = \frac{1}{a_n + 1}

By substitution, we have that the sum telescopes and reduces to

\frac{1}{a_1} - \frac{1}{a_{101}}

Since a_{101} is a very large number (the sequence has a quadratic increasing rate), we have that the integer part of the sum is


Scroll to top