+2446 Problems like these can be daunting, but it is important to perservere and be observant. I was unsure of the solution, so I decided to attempt to find the solution by drawing 0 straight lines to 5 straight lines. I have created an image of my final solutions to all of these. The red line on each indicates where I added a line from the previous image.
I will the notation \(f(n)\) to notate the maximum number of regions generated with n straight lines. I could not go much further than this, so I decided to stop here.
\(f(0)=1\\ f(1)=2\\ f(2)=4\\ f(3)=7\\ f(4)=11\\ f(5)=16\\ \)
I immediately noticed a pattern with this sequence of numbers. This is what I noticed
\(f(0)+1=f(1)\\ f(1)+2=f(2)\\ f(2)+3=f(3)\\ f(3)+4=f(4)\\ f(4)+5=f(5)\)
This is a quadratic relationship since the numbers 1,2,3,4, and 5 have a second common difference of 1. This means that there is a quadratic equation out there that represents this exact situation. Let's try and find it, shall we? Let's start this by plugging in that \(f(0)=1\):
| \(f(n)=an^2+bn+c\) | Since we already determined that the above relationship can be observed quadratically, let's plug in some values to develop that quadratic. |
| \(f(0)=1;\\ f(0)=a*0^2+b*0+c\) | This is the easiest one to simplify. |
| \(\boxed{1}\hspace{3mm}1=c\) | We now know that the constant term of the quadratic is 1. |
Let's plug in the next point, \(f(1)=2\):
| \(f(1)=2;\\ f(1)=a*1^2+b*1+c\) | Now, simplify both sides completely. |
| \(2=a+b+c\) | We established in \(\boxed{1}\) that \(c=1\), so let's substitute that in here. |
| \(2=a+b+1\) | Substract 1 from both sides. |
| \(\boxed{2}\hspace{3mm}1=a+b\) |
Let's plug in the next point, \(f(2)=4\):
| \(f(2)=4;\\ f(2)=a*2^2+b*2+c\) | Simplify from here. |
| \(4=4a+2b+c\) | As aforementioned, \(c=1\), so substitute this information in and simplify completely. |
| \(4=4a+2b+1\) | Subtract 1 from both sides again. |
| \(\boxed{3}\hspace{3mm}3=4a+2b\) | |
Our next task is to solve this system of equations. Let's do that. I will solve by substitution here because of the presence of a variable with a coefficient of one.
\(\boxed{2}\hspace{3mm}1=a+b\Rightarrow a=1-b\)
Since I have a isolated, it is possible to substitute it in for \(\boxed{3}\):
| \(\boxed{2}\hspace{3mm}a=1-b;\\ \boxed{3}\hspace{3mm}3=4(1-b)+2b\) | Solve for b in \(\boxed{3}\). |
| \(3=4-4b+2b\) | |
| \(-1=-2b\) | |
| \(b=\frac{1}{2}\) | Let's substitute this into \(\boxed{2}\). |
| \(b=\frac{1}{2};\\ \boxed{2}\hspace{3mm}a=1-\frac{1}{2}\) | |
| \(a=\frac{1}{2}\) | All the variables have been solved for now! |
After all this work, we have determined that \(f(n)=\frac{1}{2}n^2+\frac{1}{2}n+1\) is the quadratic that represents the maximum number of regions for n straight lines. In this case, n=15, so let's substitute this in and solve.
| \(n=15;\\ f(15)=\frac{1}{2}*15^2+\frac{1}{2}*15+1\) | Simplify the right hand side of the equation. |
| \(f(15)=\frac{225}{2}+\frac{15}{2}+\frac{2}{2}\) | Now, do the addition. |
| \(f(15)=\frac{242}{2}=121\text{ regions}\) | You're done! |
