Question

PLEASE HELP WITH THIS CALCULUS PROBLEM!



Suppose that you deposit $1.00 into an account that pays 100% annually.

a) How much will you have in the bank at the end of 1 year? SOLVED: $2.00

b) Suppose that the interest is compounded twice each year. How much will you have in the bank at the end of 1 year? SOLVED: $2.25

c) Calculate the amount in the bank at the end of 1 year with more compounding periods: 3, 4, 100.

d) What number does the amount in part c) seem to be approaching, as the number of compounding periods increases.

SHOW ALL WORK PLEASE!

Answer 1

check out http://www.milefoot.com/math/calculus/limits/LimitDefinitionOfE10.htm

Answer 2

Can I use that formula in Calculus, because I dont see it anywhere in my textbook

Answer 3

\[
A=A_0 \left (1+ \frac r n \right)^n
\]
Where n is the number of times it is compounded each year and e is the annual interest rate.

Answer 4

For a great explanation, please refer to this thread.
http://openstudy.com/study#/updates/4fa7037ee4b029e9dc366b88

Answer 5

for parts a, b, and c use A=A0(1+rn)n. for part d that the limit as n->\[\infty\]

Answer 6

what should I subsititue for part a and b then?

Answer 7

r not e in my post above

Answer 8

actually thats the limit as *r* goes to infinity my b

Answer 9

so for part a, r and n are both 1 right? but but what about part b?

Answer 10

You keep r the same, but you divide it by n the number of time you are compounding

Answer 11

so n would be 2?

Answer 12

what about P?

Answer 13

The limit when n is tending to Infinity of
\[
\left ( 1 + \frac 1 n\right)^n
\]
is equal to e when is approaching Infinity. This problem is
to make you believe that statement above, by asking you to compute it for n=1,2, 5, 100, etc

Answer 14

when n is approaching infinity

Answer 15

I dont understand how to do part c)

Answer 16

for parts a and b i did (1+1/1)^1 and (1+1/2)^2

Answer 17

Compute
\[
\left ( 1 + \frac 1 n \right)^n
\]
for n=3 then for n=4 and finally for n =100

Answer 18

Ok thank you

Answer 19

Ooooh, Jobo. I'm so glad you struggled so hard to get the same answer out of this guy that Alex and I gave you a half hour ago.

Answer 20

For n=50
\[\left(1+\frac{1}{50.}\right)^{
50}=2.69159

\]

Answer 21

k thanks :)

Answer 22

\[\left(1+\frac{1}{10000.}\right)
^{10000}=2.71815
\]
and \[
e\approx2.718282
\]

Answer 23

it approaches the constant e right?

Answer 24

yes

Answer 25

THANKS :)

Answer 26

yw