14. If $f(x)=(x-1)+(x-2)^{2}+\cdots+(x-10)^{10}$, then the sum of the 10 roots of $f(x)$ is ( ).
(A) 55
(B) 99
(C) 100
(D) 110
(E) 120
14.B.
The sum of the roots of an $n$-degree polynomial is equal to the opposite of the quotient of the coefficient of $x^{n-1}$ divided by the coefficient of $x^{n}$. In the original expression, the coefficient of $x^{10}$ is 1. Therefore, the sum of the roots is the opposite of the coefficient of $x^{9}$. In $(x-1)^{9}$, the coefficient of $x^{9}$ is 1, and in $(x-10)^{10}$, the coefficient of $x^{9}$ is -100. Therefore, the result is $-(1-100)=99$.