The Number of Binary Vectors Whose Sum Is Greater Than $k$ I want to determine the number of vectors $(x_1,\ldots,x_n)$, such that each $x_i$ is either $0$ or $1$ and $$\sum\limits_{i=1}^n x_i \geq k$$



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*My Approach
The number of $1$'s range from a minimum of $k$ to a maximum of $n$. Thus I must count all the ways in which I can get $$\sum\limits_{i=1}^n x_i = k,\,\,\,\, \sum\limits_{i=1}^n x_i = k +1\,\,\,\,,..., and\,\,\,\,\sum\limits_{i=1}^n x_i = n,$$
$n \choose k$ denotes all the unique ways I can select $k$ $ones$ from the $n$ possible positions.
Since the number of $1$'s range from a minimum of $k$ to a maximum of $n$, $n\choose k+1$ denotes all the unique ways I can select $k+1$ $ones$ from the $n$ possible positions.
Continuing this pattern all the way up $n$, the solution comes to 
$\sum\limits_{i=k}^n$ $n\choose i$ $=$ $n \choose k$ $+$ $n \choose k+1$ $+$ $n \choose k+2$ $+$ $\cdot \cdot \cdot +$ $n\choose n$

Please give me hints and suggestions regarding my solution.

If an approximation is good enough for your needs, you can always use the central limit theorem. Just think of each $x_i$ as a random variable, whose mean and  standard deviation would each be $\frac12$. Then, by the central limit theorem, your sum $\sum_{i=1}^nx_i$ approximately has a normal distribution with mean $\frac n2$ and standard deviation $\frac{\sqrt n}2$. So, the number of 0-1 vectors satisfying $\sum_{i=1}^nx_i>k$ would be $2^n\cdot{\rm Prob}(\sum_{i=1}^nx_i>k)$, which is approximately $$2^n\cdot\frac1{\sqrt{2\pi}\frac{\sqrt n}2}\int_k^\infty\exp\left(-\frac{2(x-\frac n2)^2}{n}\right)\,{\rm d}x.$$
In many applications, the parameter $k$ can is thought of as a function of $n$, in which case you can further approximate the integral.
