Question

n
S=∑1/(4k-3)(4k+1)
k=1
Vă rog, ajutați-mă!

Answer 1

$\Sigma \frac{1}{(4k-3)(4k+1)} = \frac{1}{4} \Sigma (\frac{1}{4k-3} - \frac{1}{4k+1}) = \\ = \frac{1}{4}( \frac{1}{1}- \frac{1}{5} + \frac{1}{5} - \frac{1}{9} + ... + \frac{1}{4n-7} - \frac{1}{4n-3} + \frac{1}{4n-3} - \frac{1}{4n+1})= \\ = \frac{1}{4} (1- \frac{1}{4n+1}).$

Observatie: la fiecare suma k porneste de la 1 si merge pana la n.