Suppose [x] denote the greatest integer $\le x$ and $n\in \mathbf{N}$, then 

$\underset{n\to \mathrm{\infty }}{lim} \frac{\left[{ }^n{C}_o{x}^2\right]+\left[{ }^n{C}_1{x}^2\right]+\cdots +\left[{ }^n{C}_n{x}^2\right]}{2^n-2}$ is equal to

We know $x-1<\left[x\right]\le x\mathrm{\forall }x\in \mathbf{R}$, there fore ${ }^n{C}_k{x}^2-1<\left[{ }^n{C}_k{x}^2\right]{\le }^n{C}_k{x}^2$

$\begin{array}{l}\Rightarrow \sum _{k=0}^n \left({ }^n{C}_k{x}^2-1\right)<\sum _{k=0}^n \left[{ }^n{C}_k{x}^2\right]\le \sum _{k=0}^n{ }^n{C}_k{x}^2\\ \Rightarrow \frac{x^2\left(2^n\right)-(n+1)}{2^n-2}<\frac{1}{2^n-2}\sum _{k=0}^n \left[{ }^n{C}_k{x}^2\right]\\ \le \frac{x^2\left(2^n\right)}{2^n-2}\end{array}$

Taking limit as $n\to \mathrm{\infty }$ ,and using sandwich theorem, we get

$\underset{n\to \mathrm{\infty }}{lim} \frac{1}{2^n-2}\sum _{k=0}^n \left[{ }^n{C}_k{x}^2\right]=x^2$