Series Demystified-1

On 'demystifying' the summation, we get

$\displaystyle \sum_{n=1}^{50} \dfrac{n^2 + 3n + 1}{n^2 + 3n + 2} = \sum_{n=1}^{50} 1 - \dfrac{1}{(n+1)(n+2)}$

which further can be identified to be a telescopic series

$\displaystyle \sum_{n=1}^{50} 1- \left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right)$

On expanding, we get:

$\displaystyle 50 - \left( \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots + \dfrac{1}{51} - \dfrac{1}{52} \right)$

$\Rightarrow \displaystyle 50 - \left( \dfrac{1}{2} - \dfrac{1}{52} \right)$

$\Rightarrow \displaystyle 50 - \dfrac{50}{104} = \dfrac{2575}{52}$

$A = 2575 \text{ \& } B = 52 \Rightarrow A+B = 2627$

Simple telescopic series.