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

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?

• Also, since the question does not specify "compounded every X days/months/quarters", it must be simple interest. – RonJohn Nov 17 '20 at 2:54
• The exercise lacks necessary information. Where is the interest paid? To the depositor? Into the account? – DJohnM Nov 17 '20 at 7:02
• Hi DJohnM, the problem is that there is no other information. The question as I wrote it is all I have to go on...@DJohnM – Ssandro_99 Nov 17 '20 at 12:29
• Given the question as it stands, what would be the most "likely" solution? @DJohnM – Ssandro_99 Nov 17 '20 at 12:32
• Your revised solution is nearly correct if the ha'penny is still a monetary unit that banks use. Else that 1278.225 would get rounded off to 1278.23 or 1278.22 depending on which convention the bank follows. There once was a bank programmer who rounded the interest paid into all accounts except his own downwards, and put the sum of the fractional parts of each penny into his own account as interest paid out. It passed all the audit tests, e.g. 3.5% of total deposits (which is what should have been paid out as interest) equals sum of all interest payments deposited into individual accounts. – Dilip Sarwate Nov 17 '20 at 13:05

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

$s=\sum_{k=1}^{n}d(1+r)^k=\frac{d(1+r)((1+r)^k-1)}{r}$

``````d = 200
r = 0.035
n = 3

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

``````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
``````
• Hi, your final version for b[n] can't correct, since it can be negative (for example when n=3) @ChrisDegnen – Ssandro_99 Nov 19 '20 at 12:14
• @Ssandro_99 I don't understand what you mean. Can you expand your example please? – Chris Degnen Nov 19 '20 at 14:36