1.

If `t_n=1/4(n+2)(n+3)` for `n=1, 2 ,3,....` then `1/t_1+1/t_2+1/t_3+....+1/(t_(2003))=`A. `(4040)/(6063)`B. `(4040)/(6069)`C. `(8080)/(6065)`D. `(8080)/(6069)`

Answer» Correct Answer - D
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)`


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