Solution:

The answer is (C). Let us denote by $S$ the sum of the 100 terms of the sequence. We have:
$$\frac{S}{2}=\frac{1}{2}\left(a_{1}+a_{2}+\ldots+a_{99}+a_{100}\right)=\frac{a_{1}+a_{2}}{2}+\frac{a_{3}+a_{4}}{2}+\ldots+\frac{a_{99}+a_{100}}{2}.$$
We observe that each of the addends on the rightmost side is the arithmetic mean of two consecutive terms of the sequence, which we know to be equal to the index of the second term. We thus obtain
$$\frac{S}{2}=2+4+\ldots+100=2(1+2+\ldots+50)$$
and recalling that the sum of the first $n$ integers equals $\frac{n(n+1)}{2}$ we conclude that
$$$$
$\text{S=4(1+2+ +50)=4 50}$ 51}{2}=5100.