Am I using the right method for this compound / simple interest problem with annual deposits?

Question

I am trying to solve the following exercise:
A bank account is opened at the beginning of the year 2013 by making a one-off deposit of £1000. A further deposit of £200 is made at the beginning of each subsequent year. The rate of interest is 3.5% and the interest is paid at the end of each year. Find the total amount in the account at the end of the year 2023, just after the interest is paid.
What I don't understand is whether interest is simple or compounded. My approach I was initially taking is via compounding. i.e.
At the end of year 1 we have £1000(1+0.035) = £1035
At the beginning of year 2 we have £1035 + £200 = £1235 (due to £200 deposit)
At the end of year 2 we then have £1235(1+0.035) = £1278.225
At the beginning of year 3 we have £1278.225 + £200 = £1478.225 (another £200 deposit)
At the end of year 3 we have £1478.225(1+0.035) = £1529.96
And so on. And then obviously one can derive the formula for an arbitrary year n. Is this correct?

Chris Degnen · Answer

A recurrence expression for the balance b in year n
b[n + 1] = (b[n] + d) (1 + r)

where b[0] = a - d

resolves to
b[n] = (((r + 1)^n) (a r + d) - d (r + 1))/r

So with  a =  1000
         d =   200
         r = 0.035

e.g.  b[3] = 1529.96

This works for all n except the initial condition b[0]
Derivation using Wolfram Alpha with input in Mathematica format.

Alternatively, putting the solution together in parts.
First the recurring deposits, by induction

d = 200
r = 0.035
n = 3

∴ s = (d (r + 1) ((r + 1)^n - 1))/r = 642.99

And adding the initial deposit
a = 1000;
b = (a - d) (1 + r)^n + s = 1529.96

In general
b(n) = (a - d) (1 + r)^n   +   (d (r + 1) ((r + 1)^n - 1))/r

However, this still has the problem with the initial condition where n = 0.
So
 b[n] = (((r + 1)^n) (a r + d) - d (r + 1))/r    where   n > 0

b[0] = a

Am I using the right method for this compound / simple interest problem with annual deposits?

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