3. Compute the sum

$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3 n-2)(3 n+1)}
$$
3. First solution. We will prove by mathematical induction that the equality

$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}
$$

a) For $n=1$, the equality holds.

b) Suppose this equality holds for a natural number $k$, i.e., suppose

$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3 k-2)(3 k+1)}=\frac{k}{3 k+1}
$$

We will show that it will also hold for $n=k+1$.

$$
\begin{aligned}
& \quad \text { Indeed, } \\
& \frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3 k-2)(3 k+1)}+ \\
& +\frac{1}{(3 k+1)(3 k+4)}=\frac{k}{3 k+1}+\frac{1}{(3 k+1)(3 k+4)}= \\
& =\frac{3 k^{2}+4 k+1}{(3 k+1)(3 k+4)}=\frac{(3 k+1)(k+1)}{(3 k+1)(3 k+4)}=\frac{k+1}{3 k+4} .
\end{aligned}
$$

Thus, the validity of the equality is proven for all $n$.

Second solution. We will use the formula

$$
\frac{1}{(3 n-2)(3 n+1)}=\frac{1}{3(3 n-2)}+\frac{1}{3(3 n+1)} \text {. }
$$

Applying it to all terms of our sum, we get

$$
\begin{aligned}
& \frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3 n-5)(3 n-2)}+ \\
& \quad+\frac{1}{(3 n-2)(3 n+1)}=\left(\frac{1}{3 \cdot 1}-\frac{1}{3 \cdot 4}\right)+ \\
& +\left(\frac{1}{3 \cdot 4}-\frac{1}{3 \cdot 7}\right)+\left(\frac{1}{3 \cdot 7}-\frac{1}{3 \cdot 10}\right)+\ldots+ \\
& \quad+\left(\frac{1}{3(3 n-5)}+\frac{1}{3(3 n-2)}\right)+ \\
& \quad+\left(\frac{1}{3(3 n-2)}+\frac{1}{3(3 n+1)}\right)
\end{aligned}
$$

After expanding the brackets, all terms except the first and the last will cancel each other out, and we will get

$$
\begin{aligned}
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10} & +\ldots+\frac{1}{(3 n-2)(3 n+1)}= \\
& =\frac{1}{3}-\frac{1}{3(3 n+1)}=\frac{n}{3 n+1} .
\end{aligned}
$$

The equality is proven.