Three, (50 points) Among all the coefficients in the expansion of $(1+x)^{2015}$, how many are multiples of 7?
Three, generalize the problem.
First, prove a lemma.
Lemma Let $m$ be a positive integer, $p$ a prime, $p \leqslant m$. If in the $p$-ary representation of $m$, there are $k_{1}$ ones, $k_{2}$ twos, ..., $k_{p-1}$ $(p-1)$s, then in the coefficients of the expansion of $(1+x)^{m}$, the number of multiples of $p$ is
$$
m+1-2^{k_{1}} 3^{k_{2}} \ldots p^{k_{p-1}}.
$$

Proof First, find the number of elements in the set
$$
M=\left\{\mathrm{C}_{m}^{n} \mid n=0,1, \cdots, m\right\}
$$

that are coprime with $p$.
Let $m=n+k$.
Since $p$ is a prime and $p \nmid \mathrm{C}_{m}^{n}$, we have $\left(p, \mathrm{C}_{m}^{n}\right)=1$.
And $\mathrm{C}_{m}^{n}=\frac{m!}{n!\cdot k!}=\frac{(n+k)!}{n!\cdot k!}$, so
$$
\sum_{i \geqslant 1}\left[\frac{n+k}{p^{i}}\right]=\sum_{i \geqslant 1}\left[\frac{n}{p^{i}}\right]+\sum_{i \geqslant 1}\left[\frac{k}{p^{i}}\right],
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Notice that for each $i$ we have
$$
\begin{array}{l}
{\left[\frac{n+k}{p^{i}}\right] \geqslant\left[\frac{n}{p^{i}}\right]+\left[\frac{k}{p^{i}}\right] .} \\
\text { Hence }\left[\frac{n+k}{p^{i}}\right]=\left[\frac{n}{p^{i}}\right]+\left[\frac{k}{p^{i}}\right](i=1,2, \cdots) .
\end{array}
$$

Let $m=n+k=a_{0}+a_{1} p+a_{2} p^{2}+\cdots+a_{r} p^{r}$, $n=b_{0}+b_{1} p+b_{2} p^{2}+\cdots+b_{r} p^{r}$,
$$
k=c_{0}+c_{1} p+c_{2} p^{2}+\cdots+c_{r} p^{\prime},
$$

where $0 \leqslant a_{i}, b_{i}, c_{i} \leqslant p-1(i=0,1, \cdots, r)$.
From equation (1) we get
$$
\begin{array}{l}
a_{r}=b_{r}+c_{r}, a_{r-1}=b_{r-1}+c_{r-1}, \cdots, \\
a_{0}=b_{0}+c_{0} .
\end{array}
$$

For each equation $a_{i}=b_{i}+c_{i}$, $b_{i}$ can take $0, 1, \cdots, a_{i}$, a total of $a_{i}+1$ values, so in the set $M$, there are $\left(a_{0}+1\right)\left(a_{1}+1\right) \cdots\left(a_{r}+1\right)$ values of $n$ such that $p \nmid \mathrm{C}_{m}^{n}$.

If the $p$-ary representation of $m$ is $m=\left(a_{r} a_{r-1} \cdots a_{1} a_{0}\right)_{p}$, and the digits $a_{0}, a_{1}, \cdots, a_{r}$ contain $k_{1}$ ones, $k_{2}$ twos, ..., $k_{p-1}$ $(p-1)$s, then
$$
\left(a_{0}+1\right)\left(a_{1}+1\right) \cdots\left(a_{r}+1\right)=2^{k_{1}} 3^{k_{2}} \cdots p^{k_{p-1}},
$$

which means there are $2^{k_{1}} 3^{k_{2}} \cdots p^{k_{p-1}}$ numbers in the set $M$ that are coprime with $p$.
Thus, in the set $M$, there are $m+1-2^{k_{1}} 3^{k_{2}} \ldots p^{k_{p-1}}$ numbers that are multiples of $p$.
Returning to the original problem.
Since the septenary representation of 2015 is $(5606)_{7}$, and its digits contain one 5 and two 6s, then in the coefficients of the expansion of $(1+x)^{2015}$, the number of multiples of 7 is
$$
2016-6 \times 7^{2}=1722 \text { (numbers). }
$$