Throwing coins until get 2 heads I have coin, and want to get 2 heads exactly. I will throw it until this condition is met. 
What is expected number of tries for this condition? 
I know that it would be $$\sum\limits_{n=2}^\infty P(X=n)n=0.5^n \cdot n\cdot(n-1)$$
however I don't have an idea how to solve that sum because we didn't learn how to.
I have knowledge just how to solve geometrical and arithmetical progression's sum.
Maybe it's possible to get expected value using Poisson distribution?
 A: Your analysis is right. The probability that $X=n$ is indeed $(n-1)(0.5)^n$.  This is because we need to have exactly $1$ head in the first $n-1$ tosses (probability $(n-1)(0.5)^{n-1}$) and then a head (probability $0.5$).  So the required expectation is
$$\sum_{n=2}^\infty (n)(n-1)(0.5)^n.  \qquad (\ast) $$
To get a closed form for this, note that (if $|x|<1$) then
$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\cdots.$$
Differentiate both sides with respect to $x$, twice. We get
$$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+5x^4+\cdots,$$
and then
$$\frac{2}{(1-x)^3}=2+(3)(2)x+(4)(3)x^2+(5)(4)x^3+ \cdots.$$
Put $x=0.5$.
We get
$$16=2+(3)(2)(0.5)^1 +(4)(3)(0.5)^2+ (5)(4)(0.5)^3+\cdots. \qquad(\ast\ast)$$
To make the right-hand side of $(\ast\ast)$ equal to $(\ast)$, we need to multiply by $(0.5)^2$. So our expectation is $(16)(0.5)^2$, which is $4$.
There are better (meaning more probabilistic) ways of tackling the problem.  For example, let $X_1$ be the waiting time, that is, number of tosses, until the first success (head), and let $X_2$ be the waiting time between the first success and the second. Then $X=X_1+X_2$, and therefore $E(X)=E(X_1)+E(X_2)$.  The random variables $X_1$ and $X_2$ each have geometric distribution with parameter $p=1/2$.  The expectation of a geometric distribution with parameter $p \ne 0$ may be something you have already seen: it is $\frac{1}{p}$.  Note that the same idea works with essentially no change if $X$ is the number of tosses until, say, the $17$-th head.
A: The probability that it will take exactly $n$ throws is the probability of getting exactly one head in the first $n-1$ throws, times the probability of getting heads on the $n$th throw, so it's $((n-1)/2^{n-1})(1/2)$, which is $(n-1)/2^n$. So the expected number of throws is $\sum_{n=2}^{\infty}n(n-1)/2^n$. Series similar to this one have come up many times on this site. See, for example, Generalizing $\sum \limits_{n=1}^{\infty }n^{2}/x^{n}$ to $\sum \limits_{n=1}^{\infty }n^{p}/x^{n}$
