6. The sequence $\left\{a_{n}\right\}$ is called a difference-ratio sequence if and only if the sequence satisfies $a_{0}=0,\left\{a_{n+1}-q a_{n}\right\}$ forms a geometric sequence with a common ratio of $q$. $q$ is called the difference-ratio of this difference-ratio sequence. Then, when the difference-ratio is greater than 1 and the sequence is formed by natural numbers within 100, the maximum number of terms is ( . )
(A) 4
(B) 5
(C) 6
(D) 7
6.C.

Let the $n+1$ terms of the geometric sequence composed of natural numbers within 100 be
$$
a_{k}=k a_{1} q^{k-1} \quad(k=0,1,2, \cdots, n) \text { and } q>1 \text {. }
$$

From $\frac{a_{k+1}}{a_{k}}=\frac{k+1}{k} \cdot q$, we know that $q$ is a positive rational number, so we can let $q=$ $\frac{p}{r}$ (where $p, r \in \mathbf{N}_{+}, (p, r)=1$ and $p \geqslant r+1$).
When $r=1$, then from
$$
100 \geqslant a_{n}=n a_{1} q^{n-1} \geqslant n q^{n-1}=n p^{n-1} \geqslant n 2^{n-1},
$$

we know that $n \leqslant 5$. When $a_{k}=k 2^{k-1}(k=0,1,2,3,4,5)$, there are exactly 6 terms that meet the conditions.

When $r \geqslant 2$, if $n \geqslant 6$, from $r^{k-1} \mid k a_{1}(k=1,2, \cdots, n)$, we know $r^{5} \mid 16 a_{1}$, thus, $100>a_{1} \geqslant \frac{r^{5}}{6}$. Hence $r \leqslant 3$.
If $r=3$, then $3^{4} \mid a_{1}$, so $a_{1}=81$.
From $a_{n}=n a_{1}\left(\frac{p}{3}\right)^{n-1} \geqslant 81 n\left(\frac{4}{3}\right)^{n-1}$, we know that $n \leqslant 1$.
If $r=2$, then $2^{4} \mid a_{1}, a_{1} \geqslant 16$ and
$$
100 \geqslant a_{n}=n a_{1}\left(\frac{p}{2}\right)^{n-1} \geqslant 16 n\left(\frac{3}{2}\right)^{n-1} \text {. }
$$

Thus, $n \leqslant 2$, which is a contradiction.
In summary, the maximum value of $n$ is 6.