Question

can someone please help me with this
a) Calculate the final amount of a 20-year long-term savings plan if a payment of
$2000 is made at the end of every year. Interest is 8.5% compounded quarterly.
b) What happens to the final value of the long-term savings plan if a payment is
made at the beginning of every year instead of at the end of every year?

Answer 1

let's derive the formula by manually computing first years
1 st Year = D
\[2\text{nd} \text{Year}= D\left(1+\frac{R}{4}\right)^4+D= D\left(\left(1+\frac{R}{4}\right)^4+1\right)\]
\[3\text{rd} \text{Year}= D\left(\left(1+\frac{R}{4}\right)^4+1\right)\left(1+\frac{R}{4}\right)^4+D= D\left(\left(1+\frac{R}{4}\right)^{4*2}+\left(1+\frac{R}{4}\right)^4+1\right)\]
\[4\text{th} \text{year}=\left(D\left(\left(1+\frac{R}{4}\right)^{4*2}+\left(1+\frac{R}{4}\right)^4+1\right)\right)\left(1+\frac{R}{4}\right)^4+D=D\left(\left(1+\frac{R}{4}\right)^{4*3}+\left(1+\frac{R}{4}\right)^{4*2}+\left(1+\frac{R}{4}\right)^4+1\right)\]

Answer 2

\[\text{Nth} \text{year}=\text{ }D\left(1+\left(1+\frac{R}{4}\right)^4+\left(1+\frac{R}{4}\right)^{4*2}+\left(1+\frac{R}{4}\right)^{4*3}+\text{...}\text{...}+\left(1+\frac{R}{4}\right)^{4*(n-1)}\right)\]
Rewritten using Geometric Sum Formula
\[D\left(\frac{1-\left(\left(1+\frac{R}{4}\right)^4\right)^N}{1-\left(1+\frac{R}{4}\right)^4}\right)\]

Answer 3

Plug in Numbers
\[2000\left(\frac{1-\left(\left(1+\frac{.085}{4}\right)^4\right)^{20}}{1-\left(1+\frac{.085}{4}\right)^4}\right)\]=99770.2

Answer 4

yes this is correct

Answer 5

if it was made at beginning of each year,
the difference in the formula would be that the first term in the sequence would be (1+.085/4)^4 not 1 and each consecutive terms exponent would be shifted up 1
=108,524.9

Answer 6

if you are able, financial calculators are useful for these type of problems to check your work...i use a TI BAII

Answer 7

thanks buddy