The formula you are using assumes a deposit at end of each year;

but with your problem, the deposit is made at beginning of each year.

Also, the "i" used in formula must be the annual "i", to coincide with the deposit frequency.

This means i = (1 + .11/4)^4 - 1 =~ .11462, or ~11.462%;

makes sense that an annual rate paid more frequently than annual ends up higher, right?

The formula for Future Value of deposits made at beginning of period is:

F = D * {[(1 + i)^(n + 1) - 1] / i - 1}

So, to calculate D:

D = F / {[(1 + i)^(n + 1) - 1] / i - 1} : see NOTE below

D = deposit required (?)

F = future value (50000)

n = number of periods (3)

i = interest per period [(1 + .11/4)^4 - 1] = .11462

D = 50000 / [(1.11462^4 - 1) / .11462 - 1] = 13,362.60

And your "account" will look like:

Code:

DATE DEPOSIT INTEREST BALANCE
Jan.01/1 13,362.60 .00 13,362.60
Jan.01/2 13,362.60 1,531.64 28,256.84
Jan.01/3 13,362.60 3,238.84 44,858.28
Dec.31/3 5,141.72 50,000.00

NOTE:

Can be simplified to:

D = F * i / [(1 + i) * ((1 + i)^n - 1)]

Usually a good idea to make (1 + i) a variable, like let r = 1 + i, then equation easier to handle:

D = F * i / [r * (r^n - 1)] where r = 1 + i