TITLE: A possible theorem
QUESTION [5 upvotes]: So i was playing around with members of a random power set, and i came to a revelation(at least to me it was). Say $A=\{1,2,3\}$ then for arbitrary $k,n\in Z^+$, $n=|A|$ and $\mathscr{P}(A)=\{\varnothing,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$, $|\mathscr{P}(A)|=2^n.$ There are ${n\choose n-1}$ ways of choosing ($n-1$)- element sets in $\mathscr{P}(A)$  so that's ${3\choose1}+{3\choose2}+{3\choose3}=7$ so we're short by one, playing around once more and we have    $2{n\choose n}+{n\choose n-1}+{n\choose n-2}=2^n$ and by generalizing:
$2{n\choose n}+{n\choose k}+...+{n\choose k-(n-2)}=2^n$ for $1\le k\le n-1$. 
Pardon my ignorance, but is there a theorem for this(if the equation is correct), and is this how research in math is done(if it is, then it must be pretty darn exciting!!!)

REPLY [5 votes]: The sums you suggest are not $2^n$ (maybe check a few simple cases?), the actual identity being
$$
\sum_{k=0}^n{n\choose k}=2^n.
$$
This is a case of the binomial theorem, which states that
$$
\sum_{k=0}^n{n\choose k}x^k=(x+1)^n,
$$ 
and, more generally, that
$$
\sum_{k=0}^n{n\choose k}x^ky^{n-k}=(x+y)^n.
$$
Happy reading!