Use PMT equals StartFraction Upper P (StartFraction r Over n EndFraction )Over [1 minus (1 plus StartFraction r Over n EndFracti
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The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for 
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
, and substituting this into the recurrence, you find that 
for all 
.
Next, the linear term tells you that
, or 
.
Now, if
is the first term in the sequence, then by the recurrence you have
and so on, such that
for all 
.
Finally, the quadratic term gives
, or 
. Then by the recurrence,
and so on, such that
for all.
Now, the solution was proposed to be
so the general solution would be
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
Next, the linear term tells you that
Now, if
and so on, such that
Finally, the quadratic term gives
and so on, such that
for all
Now, the solution was proposed to be
so the general solution would be