Question

cum se rezolva: 1/2+1/6+1/12+1/20+1/30+...+1/x(x+1)=2011/2012, unde x apartineQ*

Answer 1

[tex]\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x(x+1)}=\frac{2011}{2012}\\ \frac{1}{2}+\frac{1}{2*3}+\frac{1}{3*4}+\frac{1}{4*5}+.....+\frac{1}{x(x+1)}=\frac{2011}{2012}\\ \frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2012}\\ 1-\frac{1}{x+1}=\frac{2011}{2012}\\ \frac{x}{x+1}=\frac{2011}{2012}\\ 2012x=2011x+2011\\ x=2011[/tex]