Question

cum fac acest exercitiu?

Answer 1

   
[tex]\displaystyle\\ \left(1+ \frac{1 }{2}\right)\cdot \left(1+ \frac{1 }{3}\right)\cdot \left(1+ \frac{1 }{4}\right)\cdot ...\cdot \left(1+ \frac{1 }{99}\right)\cdot \left(1+ \frac{1 }{100}\right)=\\\\\\ =\left(\frac{2}{2}+\frac{1 }{2}\right)\cdot \left(\frac{3}{3}+ \frac{1 }{3}\right)\cdot \left(\frac{4}{4}+ \frac{1 }{4}\right)\cdot ...\cdot \left(\frac{99}{99}+ \frac{1 }{99}\right)\cdot \left(\frac{100}{100}+ \frac{1 }{100}\right)= [/tex]

[tex]\displaystyle\\ =\frac{3 }{2}\cdot \frac{4 }{3}\cdot \frac{5 }{4}\cdot ...\cdot \frac{100}{99} \cdot \frac{101}{100}=\\\\ \text{(Se simplifica numaratorul de la fractia precedenta cu numitorul}\\ \text{de la fractia urmatoare.)}\\\\ =\frac{1 }{2}\cdot \frac{101}{1}= \boxed{\bf \frac{101}{2}} [/tex]