2022/052) Simplify expression $(\frac{1}{2^3-1} + \frac{1}{2}) (\frac{1}{3^3-1} + \frac{1}{2}) \cdots (\frac{1}{100^3-1} + \frac{1}{2})$

We have let $t_n =   \frac{1}{t^3-1} + \frac{1}{2}$

$= \frac{1}{2}\frac{t^3+1}{t^3-1}$

=$\frac{1}{2}\frac{(t+1)(t^2-t+1)}{(t-1)(t^2+t+1)}$

Now given value = $\prod_{n=2}^{100} \frac{1}{2}\frac{(t+1)(t^2-t+1)}{(t-1)(t^2+t+1)}$

$= \frac{1}{2^{99}} \prod_{n=2}^{100} \frac{(t+1)}{(t-1)}\frac{(t^2-t+1)}{(t^2+t+1)}$

now let us compute the 2 products 

$\prod_{n=2}^{100} \frac{(t+1)}{(t-1)}= \frac{3}{1}.\frac{4}{2}.\frac{5}{3}\cdots \frac{98}{96}. \frac{99}{97}. \frac{100}{98}. \frac{101}{99}$

In the above product the numerator of $n^{th}$ term is same as denoniator of $n+2^{nd}$ term giving

$\prod_{n=2}^{100} \frac{(t+1)}{(t-1)} = \frac{100 * 101}{1 * 2}\cdots(1)$

let us compute 2nd product

$\prod_{n=2}^{100} \frac{(t^2-t+1)}{(t^2+t+1)}= \frac{3}{7} ( \frac{ 7}{13} \cdots \frac{9901}{10101}$

In the above product the numerator of $n^{th}$ term is same as denoniator of $n-1^{st}$ term giving

$\prod_{n=2}^{100} \frac{(t^2-t+1)}{(t^2+t+1)}= \frac{3}{10101}\cdots(2)$

using (1) and (2) we get the required result

$= \frac{1}{2^{99}}\frac{3}{10101} * 5050= \frac{2525}{2^{98} * 3367}$