[$$\frac{1}{1\times2}$$+$$\frac{1}{2\times3}$$+$$\frac{1}{3\times4}$$+........$$\frac{1}{99\times100}$$] is equal to

A

$$\frac{1}{9900}$$

B

$$\frac{99}{100}$$

C

$$\frac{100}{99}$$

D

$$\frac{1000}{99}$$