Question

[Văn Thăng Nguyễn]

\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).\) ... . \(\left(\dfrac{1}{99}+1\right)\)

Answer 1

[boi đz]

\(\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot.....\cdot99}\\ =\dfrac{100}{2}=50\)

Answer 2

[Phạm Minh Quân]

=3/2.4/3.5/4.....100/99

=100/2=50