Expected number of days If it is a good day (G) there are 60% chances tomorrow will be G and 40% chances tomorrow will be bad (B). If it is a B day, there 30% chances tomorrow will be G and 70% chances tomorrow will be B. If today is B, what is the expected number of days before seeing another B?
What I think is right: $E(days)=0.7(1)+0.12(2)+0.072(3)+...$
Is this the right way? If it is, how would I sum this? If not, please let me know.
 A: Whenever we have a bad day, we count that day and stop otherwise with probability $p$ we continue to the next day. As today is a bad day, for tomorrow $p = 0.3$ of having a good day. There on if we have a good day, probability of another good day is $0.6$.
$E(X) = 1 + 0.3 \times (1 + 0.6 \times (1 + 0.6 \times(1 + ...) + ..._)$
$E(X) = 1 + 0.3 \times \frac{1}{1-0.6} = 1.75$ (using sum of infinite geometric series $ = \frac{a}{1-r} \, ,a = 1, r = 0.6)$.
If the day when the next bad day happens should not be counted in, subtract $1$.
A: 
Is this the right way? : E(days)=0.7(1)+0.12(2)+0.072(3)+...

Yes, it is!
In other words, the sequence, expressed in days, is the following
$1 \xrightarrow{\text{with probability}} p=0.7$
$2\xrightarrow{\text{with probability}} p=0.3\times0.4$
$3\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.4$
$4\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.4$
$5\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.6\times0.4$
...
$n \xrightarrow{\text{with probability}} p=0.3\times0.4\times\underbrace{0.6\times0.6\times\dots\times0.6}_{\text{(n-2) times}}$
Thus the expectation is the following
$$ \bbox[5px,border:2px solid red]
{
\mathbb{E}[X]=0.7+0.3\mathbb{E}[Y]=0.7+0.3\Bigg[1+\frac{1}{0.4}\Bigg]=1.75
\qquad (1) 
}
$$
Where $Y$ is a geometric distribution with success parameter $0.4$ and starting from $k=2$

In any case, you can solve the series in an analytical way:
Setting $0.6=p$ your series becomes
$$\mathbb{E}[X]=0.7+\frac{0.3\times0.4}{0.6}\underbrace{\sum_{x=2}^{\infty}x p^{x-1}}_{=S}$$
$$S=\sum_{x=2}^{\infty}\frac{d}{dp}p^x=\frac{d}{dp}\sum_{x=2}^{\infty}p^x=\frac{d}{dp}\frac{p^2}{1-p}=\frac{p(2-p)}{(1-p)^2}\Bigg]_{p=0.6}=5.25$$
Thus
$$\mathbb{E}[X]=0.7+\frac{0.3\times0.4}{0.6}\times 5.25=1.75$$
As alreay obtained in (1).
A: If you read well my previous answer I wrote:
In other words, the sequence, expressed in days, is the following
$1 \xrightarrow{\text{with probability}} p=0.7$
$2\xrightarrow{\text{with probability}} p=0.3\times0.4$
$3\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.4$
$4\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.4$
$5\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.6\times0.4$
...
$E(X)=0.7+0.3[2\times0.4+3\times0.4\times0.6+4\times0.4\times0.6^2+5\times0.4\times0.6^3+\dots]$
The probabilities in the [ brackets are the one of a geometric: $p(1-p)^x$ where $p=0.4$ and $(1-p)=0.6$
Is it clear now?
If the support was correctly $Y=1,2,3,4,...$ the average would have been $\frac{1}{0.4}$. Given that the support is $2,3,4,...$ instead...  you can think that your rv is
$$Z=Y+1$$
Thus for linearity of the mean you can have
$$E(Z)=1+E(Y)$$
