Verify the following formula(Euler 1755,Opera vol.X,p.280) by using $50=2\cdot 5^2=7^2+1$:
\begin{equation}
\label{eq:11.27}
\sqrt{2}=\frac{7}{5}(1+\frac{1}{100}+\frac{1\cdot 3}{100\cdot
200}+\frac{1\cdot 3\cdot 5}{100\cdot 200\cdot 300}+\hbox{etc}.)
\end{equation}
Proof:
\begin{equation}
2\cdot 5^2=7^2+1\Leftrightarrow
\sqrt{2}=\frac{1}{5}\sqrt{7^2+1}=\frac{7}{5}\sqrt{1+\frac{1}{7^2}}=\frac{7}{5}(1-\frac{1}{1+7^2})^{-\frac{1}{2}}
\end{equation}
Expande
\begin{equation}
\label{eq:1.49}
(1-x)^{\frac{-1}{2}}
\end{equation}($|x|<1$)as
\begin{equation}
\label{eq:1.50}
1+\frac{\frac{-1}{2}}{1!}(-x)+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}(-x)^2+\frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}}{3!}(-x)^3+\cdots
\end{equation}
simplify\ref{eq:1.50},we can get
\begin{equation}
\label{eq:1.54}
1+\frac{\frac{1}{2}}{1!}x+\frac{\frac{1}{2}\frac{3}{2}}{2!}x^2+\frac{\frac{1}{2}\frac{3}{2}\frac{5}{2}}{3!}x^3+\cdots
\end{equation}
Then,let $x=\frac{1}{1+7^2}=\frac{1}{50}$,\ref{eq:1.54} turned into
\begin{equation}
\label{eq:1.563434}
1+\frac{1}{100}+\frac{1\cdot 3}{100\cdot 200}+\frac{1\cdot 3\cdot
5}{100\cdot 200\cdot 300}+\cdots
\end{equation}
So\ref{eq:11.27} is verified.