Check the forward differences of the sequence.
If 
, then let 
be the sequence of first-order differences of 
. That is, for n ≥ 1,
so that 
.
Let 
be the sequence of differences of ,
and we see that this is a constant sequence, 
. In other words, is an arithmetic sequence with common difference between terms of 2. That is,
and we can solve for 
in terms of 
:
and so on down to
We solve for 
in the same way.
Then
and so on down to
<span>Probability = 0.063
Fourth try = 0.0973
Let X be the number of failed attempts at passing the test before the student passes. This
is a negative binomial or geometric variable with x â {0, 1, 2, 3, . . .}, p = P(success) = 0.7
and the number of successes to to observe r = 1. Thus the pmf is nb(x; 1, p) = (1 â’ p)
xp.
The probability P that the student passes on the third try means that there were x = 2
failed attempts or P = nb(2, ; 1, .7) = (.3)2
(.7) = 0.063 . The probability that the student
passes before the third try is that there were two or fewer failed attmpts, so P = P(X ≤
2) = nb(0, ; 1, .7) + nb(1, ; 1, .7) + nb(2, ; 1, .7) = (.3)0
(.7) + (.3)1
(.7) + (.3)2
(.7) = 0.973 .</span>
