Question

Va rog sa ma ajutati.

Answer 1

Răspuns:

Explicație pas cu pas:

a.

$\frac{3-1}{1*3}+\frac{5-3}{3*5}+...+\frac{93-91}{91*93} =1-\frac{1}{3}+ \frac{1}{3}-...+\frac{1}{91} =1+\frac{1}{91}= \frac{92}{91}$

b.

1+2+3+...+n=$\frac{n(n+1)}{2}$

1+2=2*3/2, 1+2+3=3*4/2, 1+2+3+..+30=30*31/2

avem $\frac{1}{\frac{2*3}{2} }+ \frac{1}{\frac{3*4}{2} }+...+\frac{1}{\frac{30*31}{2} }=$$\frac{2}{2*3}+\frac{2}{3*4}+...+\frac{2}{30*31} =2(\frac{1}{2*3}+\frac{1}{3*4}+...+\frac{1}{30*31} )=$

$2(\frac{3-2}{2*3}+\frac{4-3}{3*4} +...+\frac{31-30}{30*31} )=$$2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...\frac{1}{30}-\frac{1}{31})$

$=2(\frac{1}{2}-\frac{1}{31})=2(\frac{31-2}{62})=2*\frac{29}{62} =\frac{29}{31}$