Let $P(k)$ be the probability of tossing the first head on the $k$th toss. The probability of tossing the first $H$ on your $k$th toss is equivalent to the probability of tossing $(k-1)$ consecutive $T$s and then tossing a $H$ $$ P(k) = \frac{1}{2^k}. $$ So, the expected number of tosses, $N$ to toss the first $H$ is given by $$ E[N] = \sum_{k=1}^{\infty} k P(k) = \sum_{k=1}^{\infty} \frac{k}{2^k}. $$ If you recall that this infinite series is equal to 2 then we are done. Otherwise, there is a simple method to determine the value of this series. $$ E[N] = 0.5 + 2(0.5^2) + 3(0.5^3) + 4(0.5^4) + ... $$ $$ 0.5 E[N] = 0.5^2 + 2(0.5^3) + 3(0.5^4) + 4(0.5^5) + ... $$ Subtracting $E[N]$ from $0.5E[N]$ gives $$ 0.5 E[N] = 0.5 + 0.5^2 + 0.5^3 + 0.5^4 + ... = 1. $$ Therefore, $E[N] = 2$.