Expectation of total scores when rolling a die until the score is not 6 An unbiased six-sided die 1, 2, ..., 6 is used during a game, where if one player scores a 6, then he rolls the die again. The player continues playing the game until he scores another value which is NOT 6.
Find the expected value of the total score on one turn.
My working:
$\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k).$
where, $P(X=k) = \frac{1}{6}^{k-1}\frac{5}{6}$
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{2}k\frac{1}{6}^{k-1} = \frac{5}{6}(1+2(\frac{1}{6})) = \frac{10}{9}$
EDITED
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{\infty}k\frac{1}{6}^{k-1} = \frac{5}{6}\frac{1}{(1-\frac{1}{6})^{2}} = \frac{6}{5}$
But on the answer sheet, it is $\frac{21}{5}$
I would appreciate if someone can point out my mistake. Thanks!

The expected sum after $k$ rolls is not $k$ but $6(k-1)+3$ indeed if we stop at first roll it means that we get $1,2,3,4,5$ and the average sum will the mean i.e. $3$, if we rolled twice means that the first roll we get $6$ and now one of the values $1,2,3,4,5$ and so on. It means that the exact formula is 
\begin{align}
\mathbb{E}[X]&=\sum_{k=1}^{\infty}(6(k-1)+3)P(X=k)=\frac{5}{6}\sum_{k=1}^{\infty}(6(k-1)+3)(\frac{1}{6})^{k-1}=\\
& = 5\sum_{k=1}^{\infty}(k-1)(\frac{1}{6})^{k-1}+\frac{5}{2}\sum_{k=1}^{\infty}(\frac{1}{6})^{k-1}=\\
&=\frac{6}{5}+3=\frac{21}{5}
\end{align}
