Question

aratati ca: a) 1/n×(n+1)= 1/n-1/n+1, oricare ar fi n apartine N*
b) k/n×(n+k)= 1/n-1/n+k, oricare ar fi n apartine N si k apartine N*
c) fie numarul b= 5/3×8+7/8×15+9/15×24
stabiliti daca b apartine intervalului (1/4;1/3)​

Answer 1

Răspuns:

Explicație pas cu pas:

a)

$\frac{1}{n} -\frac{1}{n+1}=\frac{n+1}{n*(n+1)} -\frac{n}{(n+1)*n}=\frac{(n+1)-n}{n*(n+1)}=\frac{1}{n*(n+1)}$

b)

$\frac{1}{n} -\frac{1}{n+k}=\frac{n+k}{n*(n+k)} -\frac{n}{(n+k)*n}=\frac{(n+k)-n}{n*(n+k)}=\frac{k}{n*(n+k)}$

c)

[tex]b=\frac{5}{3*8}+ \frac{7}{8*15}+ \frac{9}{15*24}=\frac{5*5}{3*8*5}+ \frac{7}{8*15}+ \frac{3}{15*8}=\frac{25}{8*15}+ \frac{7}{8*15}+ \frac{3}{8*15}=\\\\ =\frac{25+7+3}{8*15} =\frac{35}{8*15} =\frac{7}{8*3} =\frac{7}{24} [/tex]

=>

$b=\frac{7}{24}> \frac{6}{24}=\frac{1}{4}$

$b=\frac{7}{24}< \frac{8}{24}=\frac{1}{3}$

=> b∈$(\frac{1}{4} ;\frac{1}{3})$