Let $ABC$ be a triangle. Take $n$ point lying on the side $AB$ (different from $A$ and $B$) and connect all of them by straight lines to the vertex $C$. Similarly, take $n$ points on the side $AC$ and connect them to $B$. Into how many regions is the triangle $ABC$ partitioned by these lines?

Further, take $n$ points on the side $BC$ also and join them with $A$. Assume that no three straight lines meet at a point other than $A,B$ and $C$. Into how many regions is the triangle $ABC$ partitioned now?
1. **First Set of Lines:**
   - Consider the $n$ points on side $AB$. Connecting each of these points to vertex $C$ divides the triangle into $n+1$ regions. This is because each new line from $C$ to a point on $AB$ adds one new region.

2. **Second Set of Lines:**
   - Now, consider the $n$ points on side $AC$. Connecting each of these points to vertex $B$ will intersect each of the $n$ lines from the first set exactly once. Each new line from $B$ to a point on $AC$ will intersect each of the $n$ lines from $C$ to $AB$, creating $n+1$ new regions for each line added.
   - Therefore, adding the second set of $n$ lines increases the number of regions by $n(n+1)$.
   - The total number of regions after adding the second set of lines is:
     \[
     (n+1) + n(n+1) = (n+1)^2
     \]

3. **Third Set of Lines:**
   - Now, consider the $n$ points on side $BC$. Connecting each of these points to vertex $A$ will intersect each of the $n$ lines from $C$ to $AB$ and each of the $n$ lines from $B$ to $AC$. Each new line from $A$ to a point on $BC$ will intersect each of the $2n$ lines from the previous sets, creating $2n+1$ new regions for each line added.
   - Therefore, adding the third set of $n$ lines increases the number of regions by $n(2n+1)$.
   - The total number of regions after adding the third set of lines is:
     \[
     (n+1)^2 + n(2n+1) = n^2 + 2n + 1 + 2n^2 + n = 3n^2 + 3n + 1
     \]

The final answer is $\boxed{3n^2 + 3n + 1}$.