Expert: Sherry Wallin - 6/2/2010

Question
QUESTION: Question from Kenneth:

Hello:

I have a question regarding the following compound interest and future value
calculation.

Year 1 P + rP equals balance after the first year.
Year 2 (P + rP) + r(P + rP) equals balance after the second year.
Year 3 ? equals balance after the third year.

This question is in two parts.

1. What would follow for year three?

2. I know that a pattern will develop. What will it be so that I can determine the extended pattern for following years?
(Do not use the factored version of (P + rP) + r(P + rP).)

I thank you for your reply.

ANSWER: Hi Kenneth~
    I am going to try to explain how you would develop the pattern and then simplify it for you.
P is the principal and rp is the rate times the principal so at the end of one year you would have the original principal plus the interest on the original principal or P + rP. I am going to use Po though to represent the first (initial principal) and then the principal at the end of year one will be P1 = P0 + rP0 and at the end of year two P2 = P1 + rP1, and at the end of year n, Pn = P(n-1) + rP(n-1).

What one could do and I will do is substitute back into the earlier expressions for Po:

initially:          
Po          
         

end yr 1:
Po + rPo = P1

end yr 2:
P1 + rP1 = P2
(Po + rP0) + r(Po + rP0) = P2  since P1 = Po + rPo
Notice that there are two sets of (Po + rP0) so you can factor it out as
(Po + rP0)(1 + r) = P2  ***Note there is just one factor of (1 + r) but it is for P2

end yr 3:
p2 + rP2 = P3
(P1 + rP1) + r(P1 + rP1) = P3 since P2 = P1 + rP1
simplify noting that there are two (P1 + rP1)
(P1 + rP1)(1 + r) = P3
and simplify once more replacing P1 with Po + rPo
[(Po + rPo) + r(Po + rPo)](1 + r) and notice again there are two (Po + rPo) so simplify by factoring one more
(Po + rPo)(1 + r)(1 + r) = (Po + rPo)(1 + r)^2   ***Note there is two factors of (1 + r) but it is for P3


...


end yr n:
P(n-1) + rP(n-1) = Pn
[P(n-2) + rP(n-2)] + r[P(n-2)+ rP(n-2)]

I could painstakingly work back through the numbers or you can take my word for it that at the end of year n you will have
(Po + rPo)(1 + r)^(n-1). Now since we know that we used Po to represent the original principal P and we have rewritten all the above in terms of Po we can just write Pn = (P + rP)(1 + r)^(n-1) where n = 1,2,3,...?

To convince yourself that this formula works try n = 1,2,3
n = 1 -> (P + rp)(1 + r)^(1-1) = (P + rP)(1 + r)^0 = (P + rP)(1) = P + rP   since anything to the zero power has a value of 1

n = 2 -> (P + rP)(1 + r)^(2-1) = (P + rP)(1 + r)^1 = (P + rP)(1 + r)

and

n = 3 -> (P + rP)(1 + r)^(3-1) = (P + rP)(1 + r)^2

Math Prof
        
         



---------- FOLLOW-UP ----------

QUESTION: Hello Math Prof.:

It would be easier for me if I explain what I know. P + rP equals the principle and interest for one year. I fully understand this first step.

Now r(P + rP) was added to P + rP to equal the second year. Would just keep adding r(P +rP) to the calculation produced the amount for three years P + rP + r(P + rP) + r(P + rP) and P + rP + r(P + rP) + r(P + rP) + r(P +rP) equal four years, etc. ?  I know that r(P +rP) must be added to the previous calculation, but I do not know how many.

I thank you for your follow-up reply.

Answer
Hi Kenneth~

    No, it is not as simple as adding P + rP for each additional year. For example, I have shown you that at the end of the third year you would have:(P + rP)(1 + r)^2 and if you were to use your 'theory' then it would be equal to:
P + rP + P + rP  + P + rP = 3(P + rP) which is not the same as (P + rP)(1 + r)^2. You can verify this by making up some values for P and r and calculating both representations and seeing they are not equal.

The reason it does not work the way you want it to is because at the end of the first year you have made rP on your initial P.
During the next year your 'new' principal is no longer P, but P + rP and then in the next year your new principal is
P + rP + r(P + rP) and the next year (year 3) your new principal is P + rP + r(P + rP) + r[p + rP +r(P + rP)}. You are adding an additional P + rP to each successive year but you are adding much more than that.

Maybe this is easier to see. The initial (beginning principal is P) then you earn interest of rP for the first year. Now your beginning principal for the 2nd year becomes P + rP. You will earn r*(P + rP) in addition to what you started with P + rP. Together this means you have (at the end of the 2nd year) P + rP + r(P + rP). For the 3rd year your beginning principal is
what you had at the end of the 2nd year plus r*what you had at the end of the 2nd year or
P + rP + r(P + rP) + r*[P + rP + r(P + rP)] which simplifies mathematically as (P + rP)( 1 + r) + r*[P + rP + r(P + rP)]
which simplifies further as (P + rP)(1 + r) + r*[(P + rP)(1 + r)] = (P + rP)(1 + r)(1 + r) = (P + rP)(1 + r)^2

There is a lot of algebra involved and substitutions so it is hard for a novice to follow so don't feel bad.

Math Prof