Expert: Paul Klarreich - 7/29/2009

Question
QUESTION: Mr. Desai has decided to deposit Rs. 200,000 in the bank annually. If the bank
has a policy of continuous compounding and the prevailing interest rate is 8%
how much would his deposit grow upto in 2 years?

ANSWER: Questioner: Kapil
Country: India
Category: Advanced Math
Private: No
Subject: plz reply
Question: Mr. Desai has decided to deposit Rs. 200,000 in the bank annually. If the bank
has a policy of continuous compounding and the prevailing interest rate is 8%
how much would his deposit grow upto in 2 years?
----------------------------------------------------

Annual interest rate = 0.08

Assume a principal of exactly 1.00 and a time of one year.  Then if we compound n times a year we get an amount of:

X = (1 + .08/n)^n

by the basic compound interest formula.

If we want to compound continuously, then we get:

X = lim[n->inf] (1 + .08/n)^n

Now do a basic change-of-variable:

Let n/.08 = r,  then n = .08 r, and  .08/n = 1/r
and when n --> inf,  r --> inf.

X = lim[r->inf] (1 + 1/r)^.08r

X = lim[r->inf] [(1 + 1/r)^r]^.08

X = e^.08

You will have $1 for two years and another $1 for one year.  You can handle the rest.


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

QUESTION: Dear Sir,
its fine...but please solve the entire question using the value mentioned..its very confusing...

Request you to solve the entire question and please get the final answer...

also dont leave any steps in between as i am in total soup with questions like these...

Answer
1. The first $1 grows by a factor of e^.08 each year.  So multiply:

e^.08 * e^.08

2. The second grows a factor of e^.08 in that second year.

So add: e^.08 * e^.08 + e^.08

3. Now multiply by your 200,000 whatevers.

That is it.

If you have no idea what  lim[n -> infinity] means, you are in the wrong subject.