Tổng quát

$T=\sum\limits_{k=0}^{2016}\dfrac{C^k_{2016}}{k+1}$

$\dfrac{C^k_{2016}}{k+1}=\dfrac{2016!}{(k+1)(2016-k)!k!}$
$<=>$ $\dfrac{C^k_{2016}}{k+1}=\dfrac{2016!}{(k+1)!(2016-k)!}=\dfrac{2017!}{(k+1)!(2016-k)!}.\dfrac{1}{2017}$

$<=>$ $\dfrac{C^k_{2016}}{k+1}=C^{k+1}_{2017}.\dfrac{1}{2017}$
$=>$ $T=\sum\limits_{k=0}^{2016}C^{k+1}_{2017}.\dfrac{1}{2017}$
$<=>$ $T=\dfrac{1}{2017}(C^1_{2017}+C^2_{2017}+C^3_{2017}+.....+C^{2017}_{2017})$
$<=>$ $T=\dfrac{1}{2017}(C^0_{2017}+C^1_{2017}+C^2_{2017}+C^3_{2017}+.....+C^{2017}_{2017}-C^0_{2017})$
$<=>$ $T=\dfrac{1}{2017}(2^{2017}-C^0_{2017})$

$<=>$ $T=\dfrac{2^{2017}-1}{2017}$

Vậy $T=\dfrac{2^{2017}-1}{2017}$