Question

va rog frumos, o sa gasiti toata tema in urmatoarele postari. daca nu intelegeti, souneti sa va scriu. Succes

Answer 1

S=100+1/(1x2)+100+1/(2x3)+100+1/(3x4)+100+1/(4x5)+....+100+1/(9x10)
este evident ca suma are 9 termeni
S=900+(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+.....+(1/9-1/10)
S=900+1-1/10=9009/10

Answer 2

Vom folosi urmatoarea formula:

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

Regula sirului este urmatoarea:

[tex]S=\frac{201}{1\cdot2}+\frac{601}{2\cdot3}+\frac{1201}{3\cdto4}+...+\frac{9001}{9\cdot10}\\ S=\frac{100(1\cdot2)+1}{1\cdot2}+\frac{100(2\cdot3)+1}{2\cdot3}+\frac{100(3\cdot4)+1}{3\cdot4}+...+\frac{100(9\cdot10)+1}{9\cdot10}\\ S=\underbrace{100+100+100+...+100}_{9\ \text{elemente}}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\\\\ [/tex]
[tex]S=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}\\\\ \text{Se observa ca se reduc toti termenii, mai putin primul si ultimul}\\\\ S=900+\frac{1}{1}-\frac{1}{10}=901-\frac{1}{10}=\frac{9009}{10}[/tex]