If `t_n=1/4(n+2)(n+3)forn=1,2,3, ,` then `1/(t_1)+1/(t_2)+1/(t_3)++1/(t_(2003))=` `(4006)/(3006)` b. `(3006)/(3007)` c. `(4006)/(3008)` d. `(4006)/(3009)`

Options

A. `(4040)/(6063)`
B. `(4040)/(6069)`
C. `(8080)/(6065)`
D. `(8080)/(6069)`

Answer

D

Explanation

We have,
`t_(n)=(1)/(4)(n+2)(n+3)`
`rArr" "(1)/(t_(n))=(4)/((n+2)(n+3))`
`rArr" "(1)/(t_(n))=4((1)/(n+2)-(1)/(n+3))`
`rArr" "underset(n=1)overset(2020)sum(1)/(t_(n))=4underset(n=1)overset(2020)sum((1)/(n+2)-(1)/(n+3))`
`rArr" "underset(n=1)overset(2020)sum(1)/(t_(n))=4{((1)/(3)-(1)/(4))+((1)/(4)-(1)/(5))+ . . . . +((1)/(2022)-(1)/(2023))}`
`rArr" "underset(n=1)overset(2020)sum(1)/(t_(n))=4((1)/(3)-(1)/(2023))=(2020xx4)/(3xx2023)=(8080)/(6069)`