I am not quite sure what kind of induction you are looking for, so I will just try giving a more general solution to this problem, i.e. $E[X|Y=y]$.
The point is to notice that random variable X distributed differently before and after the y-th roll, for tosses before the first one which yield 5 cannot have 5 as result, hence there are 1 out of 5 chances to get a 6. Then follow the definition of conditional expectation we have:
$E[X|Y=y]\\=\sum\limits_{x}xP\{X=x|Y=y\}\\=\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}\frac{1}{5}+0+\sum\limits_{x=y+1}^\infty x(\frac{4}{5})^{y-1}(\frac{5}{6})^{(x-y-1)}\frac{1}{6}\\=\frac{1}{5}\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}+\frac{1}{6}(\frac{4}{5})^{y-1}\sum\limits_{x=y+1}^\infty x(\frac{5}{6})^{(x-y-1)}\\=\frac{1}{5}[5^{1-y} (-5(4^y)+5^{1+y}-4^yy)-y(\frac{4}{5})^{y-1}]+\frac{1}{6}(\frac{4}{5})^{y-1}6(y+6)\\=\frac{5}{4}(4-(\frac{4}{5})^y(y+4))+(\frac{4}{5})^{y-1}(y+6)$
To quickly verify this, plug in y=1 we will get 7 and for y=5 we have 5.8192.
Note: The last line of the equation may not be obvious but the trick here is to write $S(y)=\sum\limits_{x=1}^{y}xp^x$, and notice that:
$pS(y)=\sum\limits_{x=1}^{y}xp^{x+1}=\sum\limits_{x=2}^{y+1}(x-1)p^x$
then after few steps you will deduct general formula of the given summation.