Question

S=1supra1x2+1supra2x3+1supra3x4+............+1supra49x50

Answer 1

[tex]\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\ S=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{49\cdot 50}\\ S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\\ Termenii\ se\ reduc\ si\ mai\ ramane:\\ S=1-\frac{1}{50}\\ \boxed{S=\frac{49}{50}}[/tex]