An Interesting Product

Note that $1 - \dfrac{1}{k^{2}} = \dfrac{k^{2} - 1}{k^{2}} = \dfrac{(k - 1)(k + 1)}{k*k}$ .

As a result, when we compute $\prod_{k=2}^{n} \frac{(k - 1)(k + 1)}{k*k}$ we will end up with a telescoping product, where all the "middle" terms will cancel leaving us with a product of a fraction at the beginning and a fraction at the end. To show how this works, consider the first $4$ terms of this product; we have

$\displaystyle\prod_{k=2}^{5} \dfrac{(k - 1)(k + 1)}{k*k} = \dfrac{1*3}{2*2} *\dfrac{2*4}{3*3} *\dfrac{3*5}{4*4} *\dfrac{4*6}{5*5} = \dfrac{1}{2}*\dfrac{6}{5} = \dfrac{6}{10}$ .

Notice that the final result is in the form $\frac{1}{2}*\frac{n+1}{n}$ with $n = 5$ . So since the given product goes up to $n = 100$ , the solution will be $\dfrac{1}{2} *\dfrac{101}{100} = \dfrac{101}{200}$ .

Thus $a = 101, b = 200$ and $a + b = \boxed{301}$ .

(1-1/4)(1-1/9)........(1-1/10000).

=(4-1)/4 * (9-1)/9 *...........(10000-1)/10000.

=(2^2-1)/2^2 * (3^2-1)/3^2 *................... (99^2-1)/99^2 * (100^2-1)/100^2.

=(2+1)(2-1)/2^2 * (3+1)(3-1)/3^2 ..............* (99+1)(99+10)/99^2 * (100+1)(100-1)/100^2.

= 3 * 1 * 4 * 2 * 5 * 3 * 6 * 4 .......... 99 * 97 * 100 * 98 * 101 * 99 / 2^2 * 3^2 * 4^2 * .......... 99^2 * 100^2

regrouping the terms gives us-

=1 * 2 * 3^2 * 4^2 * 5^2 * ............. 99^2 * 100 * 101 / 2^2 * 3^2 * 4^2 * .......... 99^2 * 100^2

Squares from 3 to 99 get cancelled...so,

=1 * 2 * 100 * 101 / 4 * 10000

=101/200 ... where 'a'=101, 'b'=200

hence, a+b= 101+200= 301 is the answer