Question

S=201 supra 2+601supra6+1201supra 12+2001supra20 +...+9001 supra90 
S=?

Answer 1

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