$50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$. The $50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$ after the decimal point.
$\underline{\bf{My\; Try}}::$ Let $\left(\sqrt{50}+7\right)^{50} = I+f$, where $I = $Integer part and $f = $ fractional part. and $0\leq f<1$
Now Let $\left(\sqrt{50}-7\right)^{50} = g$ and here $0\leq g<1$
Now $\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = 2\left(\binom{50}{0}\cdot (\sqrt{50})^{50}+\binom{50}{2}\cdot (\sqrt{50})^{48}\cdot 7^2+..............+\binom{50}{50}\cdot 7^{50}\right)$ 
$\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = $Integer quantity.
$\displaystyle I+f+g = $ Integer Quantity. and $0\leq f+g<2$ So $(f+g) = 0$ or $(f+g) = 1$
So $I=$Integer quantity or $I = $ Integer quantity$\;\;  -1$.
Now I did not understand how can i get it.
Help required.
Thanks
edited.

Consider the number $N=(\sqrt{50}+7)^{50}+(\sqrt{50}-7)^{50}$.
Using the Binomial Theorem, or otherwise, we can show that $N$ is an integer.
Our number $(\sqrt{50}+7)^{50}$ is a little below $N$. How much below? Courtesy of the calculator, $(\sqrt{50}-7)^{50}\approx 4\times 10^{-58}$. So the $50$-th digit after the decimal point of $(\sqrt{50}+7)^{50}$ is $9$. 
