179. The sequence $\left(a_{n}\right)$ satisfies the conditions:

$$
a_{1}=1, \quad a_{2}=5, \quad a_{n+1}=2 a_{n}-a_{n-1}(n>1)
$$

Find the formula for the general term of the sequence.
$\triangle$ The last condition can be represented in the following form:

$$
a_{n+1}-a_{n}=a_{n}-a_{n-1}
$$

This equality means that the sequence $\left(a_{n}\right)$ is an arithmetic progression. Its first term $a_{1}=1$, and the common difference

$$
d=a_{2}-a_{1}=5-1=4
$$

Using the formula for the $n$-th term of an arithmetic progression, we get:

$$
a_{n}=a_{1}+d(n-1)=1+4(n-1)=4 n-3
$$

Answer: $a_{n}=4 n-3$.