Question

Total number of regions in which '$n$' coplanar lines can divide the plane, it is known that no two lines are parallel and no three of them are concurrent, is equal to

• A. $\frac{1}{2}(n^2+n+2)$
• B. $\frac{1}{2}\left({n+3n}^2\right)$
• C. $\frac{1}{2}\left({3n+n}^2\right)$
• D. $(n^2-n+2)$

Answer 1

The correct option is A $\frac{1}{2}(n^2+n+2)$

Let number of regions for $n$ lines be $R\left(n\right)$
Clearly, $R\left(1\right)=2,R\left(2\right)=4,...$
$R\left(n\right)=R(n-1)+n$ i.e., $R\left(n\right)-R(n-1)=n$
Putting $n=2,3,....n$ and then adding, we get
$R\left(n\right)-R\left(1\right)=2+3+......+n$
$=2+2+3+4+....+n$
$=1+\frac{n(n+1)}{2}$

$=\frac{n^2+n+2}{2}$