IMC 2010 · Problem 2

Day 117th IMC · Blagoevgrad, Bulgaria

Statement

Compute the sum of the series

k=01(4k+1)(4k+2)(4k+3)(4k+4)=11234+15678+.\sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \dots.

Official solution

Hidden so you can work on the problem first.