Ta có:
`2^{2019}=(1+1)^{2019}=C_{2019} ^0+C_{2019} ^1+...+C_{2019} ^{2019}`
`1/ {k+1} C_{2018} ^k=1/{k+1} {2018!}/{k!.(2018-k)!}={2019!}/{2019.(k+1)k!(2018-k)!}={2019!}/{2019.(k+1)!(2019-(k+1))!}`
`=1/ {2019} .C_{2019} ^{k+1}`
`S=C_{2018} ^0+1/ 2 C_{2018} ^1+1/ 3 C_{2018} ^2+...+1/ {2019} C_{2018} ^{2018}`
`S=1/{0+1}C_{2018} ^0+1/ {1+1} C_{2018} ^1+1/ {2+1} C_{2018} ^2+...+1/ {2018+1} C_{2018} ^{2018}`
`S=1/{2019} .C_{2019} ^1+1/{2019} .C_{2019} ^2+1/{2019} .C_{2019} ^2+...+1/{2019} .C_{2019} ^{2019}`
`S=1/{2019} .(C_{2019} ^1+C_{2019} ^2+...+C_{2019} ^{2019})`
`S=1/{2019} .(C_{2019} ^0+C_{2019} ^1+...+C_{2019} ^{2019}-C_{2019} ^0)`
`S=1/{2019} .(2^{2019}-1)`
`S={2^{2019}-1}/{2019}`
Vậy đáp án C