Let $X : \Omega \to \mathbb{N} $ be a r.v. Then $$ \mathbb{E} \{ X \} = \sum_{n=0}^{\infty} P(X > n) $$
Can i use the expression above to calculate the expected number of trials for getting 2 consecutive heads ,I am trying to do it this way but im getting stuck.
PS I know this problem can be solved by assuming that the expected number of trials is 'x' and proceeding further.
Yes, the formula works for any random variable with non-negative integer values. If $\ X\ $ is the number of tosses of a fair coin required to obtain $2$ successive heads, then $\ P(X>n)=\frac{F_{n+2}}{2^n}\ $, where $\ F_j\ $ is the $\ j^\mathrm{\,th}\ $ Fibonacci number. Thus $$ \mathbb{E}\left\{X\right\}=\sum_\limits{n=0}^\infty \frac{F_{n+2}}{2^n} = 6\ , $$ although this certainly doesn't seem to me to be the simplest way to compute this quantity.
Answer to question below: Why is $\ P(X>n)=\frac{F_{n+2}}{2^n}\ $?
Let $\ T_n\ $ be the outcome of the $\ n^\mathrm{th}\ $ toss, $\ \pi_n = P(X>n)$, $\ \pi_0(n) = P\left(\left\{X> n\right\}\cap \left\{T_n= \mathrm{tail}\right\}\right)\ $, and $\ \pi_1(n) = P\left(\left\{X> n\right\}\cap\left\{T_n= \mathrm{head}\right\}\right)\ $. Then \begin{eqnarray} \pi_n&=& \pi_0(n)+ \pi_1(n)\\ \pi_0(n+1) &=& \left(\pi_0(n) + \pi_1(n)\right) P\left(T_{n+1}= \mathrm{tail}\right)\\ &=& \frac{\pi_n}{2}\\ \pi_1(n+1) &=& \pi_0(n)P\left(T_{n+1}= \mathrm{head}\right)\\ &=& \frac{\pi_0(n)}{2}=\frac{\pi_{n-1}}{4}\\ \pi_0(1)&=&\pi_1(1) = \frac{1}{2}\\ \pi_0(2)&=&\frac{1}{2}\\ \pi_1(2)&=&\frac{1}{4}\ . \end{eqnarray} From these equations, it follows that $\ \pi_n\ $ satisfies the recurrence $$ \pi_{n+1}=\frac{\pi_n}{2}+\frac{\pi_{n-1}}{4}\ . $$ Therefore, if we put $\ \xi_n = 2^n\pi_n $, then $\ \xi_n\ $ satisfies the Fibonacci recurrence, $$ \xi_{n+1} = \xi_n + \xi_{n-1}\ , $$ and it's also easy to check that $\ \xi_1=2\pi_1=2=F_3\ $, and $\ \ \xi_2=4\pi_2=3=F_4\ $. It follows from this that $\ \xi_n = F_{n+2}\ $ for all $\ n\ge 1\ $, and so $\ \pi_n = \frac{\xi_n}{2^n}=\frac{F_{n+2}}{2^n}\ $.