Magic cake

For this problem we have to use recursion method.
Suppose, for n lines the number of regions are $L_{n}$ .
If there is no line in the page. Then $L_{0}$ =1.
And $L_{1}$ =2,
$L_{2}$ =4=2+2= $L_{1}$ +2,
$L_{3}$ =7=4+3= $L_{2}$ +3,
$L_{4}$ =11=7+4= $L_{3}$ +4.
So, $L_{n}$ = $L_{n-1}$ +n = $L_{n-2}$ +n-1+n = $L_{0}$ +1+2+.......+n-1+n
=1+ $\frac{n(n+1)}{2}$
Thus, $L_{200}$ =1+100*201=20101

To have the maximum number of slices it's necessary that the cuts intersect 2 by 2 and don't pass through the same point.

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The $1^{st}$ cut divides the cake in 2 regions and the $2^{nd}$ cut divides both of them in 2 parts.

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The $3^{rd}$ cut, instersecting the previous 2 in 2 different points, passes through 3 regions and obtains 3 more regions.

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In general, the $n$ cut intersects the previous $n-1$ cuts and $n$ regions are divided in 2 parts.

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In general, with $n$ cuts, the number of slices is $1+(1+2+3+........+n)=1+\frac{n(n+1)}{2} \Rightarrow$ for $n=200$ we have $1+\frac{200(200+1)}{2}=20101$