Question

Ex 8 cu rezolvare multumesc

Answer 1

Ne vom folosi de urmatoarea relatie:
$\frac{1}{k(k+1)}= \frac{1+k-k}{k(k+1)}= \frac{1+k}{k(k+1)} - \frac{k}{k(k+1)}= \frac{1}{k}- \frac{1}{k+1}$

O vom aplica fiecarui termen:

$S= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}3} - \frac{1}{4} +...+ \frac{1}{n} - \frac{1}{n+1}$

Dupa cum observi, se vor reduce termenii -1/2 cu 1/2, -1/3 cu 1/3 ... -1/n cu 1/n, si vor ramane 1/1 si -1/(n+1)

 $S= \frac{1}{1}- \frac{1}{n+1}= \frac{n}{n+1} = \frac{100}{101}\rightarrow n=100$