MathFun4Kids

The Telescopic Method is a technique for calculating the summary or product of a certain series in which each term can be decomposed into multiple parts, with some of them cancelling those of the next term.

For example, to calculate the following summary, in which each term is decomposed into two parts:

$$\begin{align} S&=\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+…+\dfrac{1}{99\times 100} \\ &=\left(\dfrac{1}{1}-\cancel{\dfrac{1}{2}}\right)+\left(\cancel{\dfrac{1}{2}}-\cancel{\dfrac{1}{3}}\right)+\left(\cancel{\dfrac{1}{3}}-\cancel{\dfrac{1}{4}}\right)+…+\left(\cancel{\dfrac{1}{99}}-\dfrac{1}{100}\right) \\ &=\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{99}{100} \\ \end{align} $$

The similar technique can be used for calculating the product of a certain series, such as:

$$\begin{align} P&=\left(1-\dfrac{2}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times … \times\left(1-\dfrac{2}{97}\right)\times\left(1-\dfrac{2}{99}\right) \\ &=\dfrac{1}{\cancel{3}}\times\dfrac{\cancel{3}}{\cancel{5}}\times\dfrac{\cancel{5}}{\cancel{7}}\times … \times\dfrac{\cancel{95}}{\cancel{97}}\times\dfrac{\cancel{97}}{99}=\dfrac{1}{99} \\ \end{align} $$

Find the answers for the following questions:

1. 
   $S=\dfrac{1}{3^2-1}+\dfrac{1}{5^2-1}+\dfrac{1}{7^2-1}+…+\dfrac{1}{99^2-1}$
2. 
   $S=\dfrac{1}{1\times 2\times 3}+\dfrac{1}{2\times 3\times 4}+\dfrac{1}{3\times 4\times 5}+…+\dfrac{1}{98\times 99\times 100}$
3. 
   $S=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+…+\dfrac{1}{\sqrt{99}+\sqrt{100}}$
4. 
   $S=\dfrac{2^2+1}{2^2-1}+\dfrac{4^2+1}{4^2-1}+\dfrac{6^2+1}{6^2-1}+…+\dfrac{100^2+1}{100^2-1}$
5. 
   $S=\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times … \times\left(1-\dfrac{1}{100}\right)$
6. 
   $P=\left(1-\dfrac{1}{2^2}\right)\times\left(1-\dfrac{1}{3^2}\right)\times\left(1-\dfrac{1}{4^2}\right)\times … \times\left(1-\dfrac{1}{100^2}\right)$
7. 
   $P=\dfrac{2^3+1}{2^3-1}\times\dfrac{3^3+1}{3^3-1}\times\dfrac{4^3+1}{4^3-1}\times…\times\dfrac{100^3+1}{100^3-1}$
8. 
   $S=\sum_{n=1}^{100}\left(\dfrac{4n}{4n^4+1}\right)$
9. 
   $S=\sum_{n=1}^{100}\left(\dfrac{n^2-\dfrac{1}{2}}{n^4+\dfrac{1}{4}}\right)$
10. 
    $S=1\times 1!+2\times 2!+3\times 3!+4\times 4!+…100\times 100!$
11. 
    $S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+…+\dfrac{1}{24\sqrt{23}+23\sqrt{24}}+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}$
12. 
    $P=\dfrac{3^2-1}{3^2-4}\times\dfrac{4^2-1}{4^2-4}\times\dfrac{5^2-1}{5^2-4}\times … \times\dfrac{100^2-1}{100^2-4}$
13. 
    $P=\dfrac{2^2}{2^2-1}\times\dfrac{3^2}{3^2-1}\times\dfrac{4^2}{4^2-1}\times … \times\dfrac{50^2}{50^2-1}$

Let $a$ and $b$ are positive integers that makes $\dfrac{\sqrt{2}+\sqrt{a}}{\sqrt{3}+\sqrt{b}}$ a rational number, where $a <= b$. Find all pairs of $(a, b)$.