# Find the value of n from the sum of a math series

## Sequence and Series Math Problem: Find the value of n from the sum of a math series

How to find the value of n from the sum of the math series is given as n – 1/n

\begin{aligned} & \sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}} +...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}} = n-\frac{1}{n}\\ \end{aligned}

\begin{aligned} n=?\\ \end{aligned}

## Solution

\begin{aligned} \sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\sqrt{1+\frac{1}{1}+\frac{1}{4}}\\ \\ &=\sqrt{\frac{4+4+1}{4}}\\ \\ &=\sqrt{\frac{9}{4}}\\ \\ \sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\frac{3}{2}\\ \end{aligned}

\begin{aligned} \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}&=\sqrt{1+\frac{1}{4}+\frac{1}{9}}\\ \\ &=\sqrt{\frac{36+9+4}{36}}\\ \\ &=\sqrt{\frac{49}{36}}\\ \\ \sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\frac{7}{6}\\ \end{aligned}

\begin{aligned} \sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}&=\sqrt{1+\frac{1}{9}+\frac{1}{16}}\\ \\ &=\sqrt{\frac{144+16+9}{144}}\\ \\ &=\sqrt{\frac{169}{144}}\\ \\ \sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\frac{13}{12}\\ \end{aligned}

\begin{aligned} \sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}&=\sqrt{1+\frac{1}{9801}+\frac{1}{10000}}\\ \\ &=\sqrt{\frac{98010000+10000+9801}{98010000}}\\ \\ &=\sqrt{\frac{98029801}{98010000}}\\ \\ \sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}&=\frac{9901}{9900}\\ \end{aligned}

Now we can rearrange the series

\begin{aligned} \frac{3}{2}+\frac{7}{6}+\frac{13}{12}+....+\frac{9901}{9900}\\ \end{aligned}

Now

\begin{aligned} \frac{3}{2}+\frac{7}{6}&+\frac{13}{12}+....+\frac{9901}{9900}\\ \\ &= 1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+....+1+\frac{1}{9900}\\ \\ &= (1+1+1+...1)+\bigg(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}\bigg)\\ \\ &= 99+\bigg(1-\frac{1}{2}\bigg)+\bigg(\frac{1}{2}-\frac{1}{3}\bigg)+\bigg(\frac{1}{3}-\frac{1}{4}\bigg)+...+\bigg(\frac{1}{99}-\frac{1}{100}\bigg)\\ \\ &= 99+\bigg(1-\frac{1}{100}\bigg)\\ \\ &= 100 -\frac{1}{100}\\ \\ \end{aligned}

So

\begin{aligned} 100 -\frac{1}{100}=n -\frac{1}{n}\\ \end{aligned}

That is **n = 100**