Question

Can someone help me solve this problem?

An Initial investment of $1000 is appreciated for 8 years in an account that earns 9% interest, compounded annually. Find the amount of money in the account the end of the period.

Answer 1

to compund is to re evaluate the amount with the added interest

Answer 2

compound even

Answer 3

\[1000*(1.09)^8\]

Answer 4

Initially you have
1000

1 year later you have 9% more
1000*1.09

2 years later:
(1000*1.09)*1.09 = 1000*1.09^2

And so on.

Answer 5

A(0) = 1000
A(1) = 1000 + 1000(.09)
A(2) = 1000 + (1000 + 1000(.09)) (.09)
= 1000 + 1000(.09) + 1000(.09)(.09)
= 1000 + 1000(.09) + 1000(.09^2)
= 1000 + 1000(.09 + .09^2)
= 1000(1 + (.09 + .09^2))

A(3) = 1000(1 + (.09 + .09^2 + .09^3)) etc

Answer 6

nice amis

Answer 7

1.00 Sn = (.09 + .09^2 + .09^3 + ... + .09^(n))
-.09 Sn = ( - .09^2 - .09^3 - ... - .09^(n) - .09^(n+1))
----------------------------------------------------
1-.09 Sn = (.09 - .09^(n+1)) ; divide off the (1-.09)

Sn = (.09 - .09^(n+1))/(1-.09)

= .09 (1-.09^n)/(1-.09)

when n=8; we get .09(1-.09^8)/(1-.09) =abt. 0.0989

A(8) = 1000(1+.0989) = 1098.90

hmmm .... i missed it along the way someplace :)

the right answer is given by the fiddler:

A(8) = 1000(1+.09)^8 = 1992.56

Answer 8

http://en.wikipedia.org/wiki/Compound_interest