Question

Aflati ,,n" apartine N din egalitatea :
1 supra 2 + 1 supra 6 + 1 supra 12 + ..... 1 supra n(n+1) = 1003 supra 1004
Va rog din suflet sa ma ajutati si îmi cer scuze modul cum.am.scris dar așa am.putut scriind de pe un tel !

Answer 1

[tex]Forma\ generala:\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\ Revenind:\\ \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n(n+1)}=\frac{1003}{1004}\\ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{n(n+1)}=\frac{1003}{1004}\\ 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{n}-\frac{1}{n+1}=\frac{1003}{1004}\\ 1-\frac{1}{n+1}=\frac{1003}{1004}\\ \frac{n}{n+1}=\frac{1003}{1004}\Rightarrow \bold{n=1003}[/tex]