Question

[Nguyễn Ngọc Bích]

Mọi người giúp em với ạ. Em cần gấp!

Answer 1

[Nguyễn Hoàng Minh]

\(C=\dfrac{1}{100}-\left(\dfrac{1}{100\cdot99}+\dfrac{1}{99\cdot98}+...+\dfrac{1}{3\cdot2}+\dfrac{1}{2\cdot1}\right)\\ C=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ C=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)=\dfrac{1}{100}-\dfrac{99}{100}=-\dfrac{98}{100}=-\dfrac{49}{50}\)

Answer 2

[Minh Hiếu]

=1/100-(1/100.99 +1/99.98 +1/98.97 +...+1/3.2 +1/2.1)

=1/100 -(100-1/99+1/99+1/98+...+1/3-1/2+1/2-1)

=1/100 -(1/100-1)

=1/100+99/100

=1