For all diagonals $a$ of Pascal's triangle (figurate numbers), $\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2$? I am looking for the derivation of the closed form along any given diagonal $a$ of Pascal's triangle,
$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$
Numbered observations follow. As for the limit proposed in the title given by:
Observation 1
$$\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2,$$
when I calculate the sums numerically using MS Excel for any $a$ within the domain ($0\le a \le100$) the sum approaches 2.000000 in all cases within total steps $n\le285$. The first series with $a=0$ is a familiar geometric series, and perhaps others look familiar to you as well:
$$\sum_{k=0}^\infty {k\choose 0}\frac{1}{2^k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... =2,$$
$$\sum_{k=1}^\infty {k\choose 1}\frac{1}{2^k}=\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+\frac{5}{32}... =2,$$
$$\sum_{k=2}^\infty {k\choose 2}\frac{1}{2^k}=\frac{1}{4}+\frac{3}{8}+\frac{3}{8}+\frac{5}{16}+\frac{15}{64}... =2,$$
but it is both surprising and elegantly beautiful that these sums across all diagonals appear to approach the same value. Some additional observations from the numerically determined sums:
Observation 2
The maximum value of any term ${k\choose a}\frac{1}{2^k}$ within a diagonal $a$ for the domain $(a>0)$ is attained at $k=2a-1$ and repeated for the term immediately following ($k=2a$).
Observation 3
$$\sum_{k=a}^{2a} {k\choose a}\frac{1}{2^k}=1$$
Observation 4
$$\sum_{k=a}^{n} {k\choose a}\frac{1}{2^k} + \sum_{k=n-a}^{n} {k\choose n-a}\frac{1}{2^k}=2$$
It's very likely that the general closed form has been derived before, but searching for the past several days has produced no results. It appears that setting up the appropriate generating function may play a role, but I am at a loss as to how to proceed. Looking forward to the responses.

Using algebra
$$f_n(x)=\sum_{k=a}^n {k\choose a}x^k=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}-x^{n+1} \binom{n+1}{a} \, _2F_1(1,n+2;n+2-a;x)$$ where appears the gaussian hypergeometric function.
$$g(x)=\sum_{k=a}^\infty {k\choose a}x^k=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}$$
So
$$f_n\left(\frac{1}{2}\right)=2-\frac{ \Gamma (n+2) }{2^{n-a}\Gamma (a+1) \Gamma (n+2-a)}\,
   _2F_1\left(-a,n+1-a;n+2-a;\frac{1}{2}\right)$$
$$g\left(\frac{1}{2}\right)=2$$
Now
$$f_{2a}(x)=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}-\binom{2 a+1}{a} x^{2 a+1} \, _2F_1(1,2 a+2;a+2;x)$$
$$f_{2a}\left(\frac{1}{2}\right)=1$$
