A calculus problem by Pratik Shastri

In order to solve this problem quickly, one has to be aware of Wallis' Intergals . Being aware of these will, after some simplification, give you the answer rather easily.

1. We have to find $W_{100}$ and $W_{102}$ . The formula to find $W_n$ where $n$ is an even number ( $2p$ ) or symbolically $n = 2p$ is

$W_{2p} = \frac{(2p)!}{2^{2p}(p!)^2} \frac{\pi}{2}$

Using the formula, we simply divide $\frac{W_{100}}{W_{102}}$ to get

$E = {\frac{(100)!}{2^{100}(50!)^2}\frac{\pi}{2} \over \frac{(102)!}{2^{102}(51!)^2}\frac{\pi}{2}}$

Now, it's just matter of simplification. We immediately can see that $\frac{\pi}{2}$ will cancel out; dividing $2^{102}$ and $2^{100}$ will give us $2^2$ and $100!$ will get cancelled out by the larger factorial leaving us with $102\cdot 101$ in the denominator.

$E = {2^2 \cdot (51!)^2 \over 102\cdot 101 (50!)^2}$

Further simplification will give us

$E = \frac{102}{101}$

Therefore, our answer is 102 + 101 = 203

The top is (1/2)(3/4)(5/6)......(99/100)(pi/2). The bottom is (1/2)(3/4)(5/6)......(99/100)(101/102)(pi/2). By getting rid of like terms on the top and bottom, you end up with 1/(101/102)= 102/101. Now, 102+101=203. Just a simpler solution. :-)