[tex] \frac{201}{2}+ \frac{601}{6}+ \frac{1201}{12}+...+ \frac{9001}{90}= \frac{200+1}{2}+ \frac{600+1}{6}+ \frac{1200+1}{12}+...+ \frac{9000+1}{90}= [/tex]
[tex]100+ \frac{1}{2}+100+ \frac{1}{6}+100+ \frac{1}{12}+...+100+ \frac{1}{90} = [/tex]
[tex]100+ \frac{1}{1*2}+100+ \frac{1}{2*3} +100+ \frac{1}{3*4}+...+100+ \frac{1}{9*10} [/tex]
Observam ca suma are 9 termeni si devine:[tex]9*100+ \frac{1}{1*2}+ \frac{1}{2*3}+ \frac{1}{3*4}+...+ \frac{1}{9*10}= [/tex]
[tex]900+ \frac{1}{1}- \frac{1}{2}+ \frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4}+...+ \frac{1}{9} - \frac{1}{10}=900+1- \frac{1}{10} = [/tex]
[tex]900+ \frac{9}{10}= \frac{9009}{10}=900,9 [/tex]
Am aplicat proprietatea: [tex] \frac{1}{n(n+1)}= \frac{1}{n}- \frac{1}{n+1} [/tex]
