Question

You open a deposit account that pays 2.75% annual interest compounded monthly. There was an opening deposit of $500 and further deposits of $500 at the beginning all the following months. ?

How much money would be in the account at the end of January 20 years later?

Answer 1

`$160110.67`

Explanation:

Formula for compound interest:

`A=P(1+r/n)^(nt)`, where

`A` is the final amount,
`P` is the original amount,
`r` is the interest rate,
`n` is the number of compounding periods per year,
`t` is time in years.

We know that since the rate is the same, the multiplier of the amount would be constant albeit the amount.

Multiplier`=(1+0.0275/12)^(12xx1/12)`
`color(white)(xxxxx.)=4811/4800`

First month,

`A_1=($500)(4811/4800)^1`

Second month,

`A_2=(A_1+$500)(4811/4800)^1`
`color(white)(A_2)=(($500)(4811/4800)^1+$500)(4811/4800)^1`
`color(white)(A_2)=$500(4811/4800)^2+$500(4811/4800)^1`

From the pattern of first and second months, we can formulate a formula for this question, in which is.

`A=500(4811/4800)^(240)+500(4811/4800)^(239)+...+500(4811/4800)^1`
`color(white)(A)=500[(4811/4800)^(240)+(4811/4800)^(239)+...+(4811/4800)^1]`
`color(white)(A)=500[(4811/4800-(4811/4800)^241)/(1-4811/4800)]`
`color(white)(A)=160110.67`

There would be `$160110.67` in the account after `20` years.

Props to Mr.Mike!

Answer 2

Suppose that only one deposit of $500 was made. Then we have:

The total sum in the account after 20 years is approximately
`$866.08` to 2 decimal places.

Explanation:

Note that we are using `r% -> rxx1/100=2.75/100`

Given: `A=P(1+r/(nxx100))^(nt)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In the account at the end of each calculation cycle

Month 1 `->P(1+r/(1200))`

2nd month includes the first `->P(1+r/1200)(1+r/1200)`

3rd month includes all the previous

`->P(1+r/1200) (1+r/1200)(1+r/1200)`

So each month we gain an extra `(1+r/1200)` and the whole giving the amount in the account at that time.

`color(brown)("This is exponential growth not a series. ")`

If you use a series type of approach your are saying that there is more in the account than is actually in it. (cooking the books)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are 12 calculation cycles in 1 year so the exponent is:
`"count of cycles per years" xx" count of years"-> nt=12xx20`

`=>A=$500(1+2.75/(1200))^(12xx20)`

`A=$500(1202.75/1200)^240`

Take logs of both sides

`log(A)=log(500)+240[color(white)(2/2)log(1202.75)-log(1200)color(white)(2)]`

`A=$866.08135.....`

`A~~$866.08` to 2 decimal places.

Answer 3

Suppose a deposit of $500 was made at the beginning of every month for 20 years.

`s=$160110.67` to 2 decimal places

Explanation:

`"1st month "->color(red)( $500(1202.75/(1200)))`

Set `1202.75/1200` as r

`"2nd month "->color(green)( [color(red)(500r^2)+500r]`

`"3nd month "->color(green)( [color(red)(500r^3+500r^2)+500r])`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given that the full term is 20 years =240 calculation cycles we end up with:

`"240th month "->color(green)( [color(red)( 500r^240+500r^239+500r^239+...+500r )])`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the general case for `n` months and the monthly deposit of `P`

`->Pr^n+Pr^(n-1)+Pr^(n-2)+...+Pr`

Factor out the `P` and set the sum as `s`

`s=P(r^n+r^(n-1)+r^(n-2)+...+r)" ".................Equation(1)`

Multiply both sides by `r`

`sr=P(r^(n+1)+r^n+r^(n-1)+...+r^2)" ".............Equation(2)`

`Eqn(1)-Eqn(2)`

`s-sr=P(r-r^(n+1))`

`s=(P(r-r^(n+1)))/(1-r)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting all of this together

`s=(500[1202.75/1200-(1202.75/1200)^(241)])/(1-1202.75/1200)`

`s=$160110.6740...`

`s=$160110.67` to 2 decimal places