Posted by visoth on Thursday, January 31, 2013 at 10:21am.
(1-u)^n / (1-u) = 1+u+u^2+...+u^n-1
That help?
im assuming you replaced (1+i) with u right if so i thinki kinda get
r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + r(1+i)^-n
looks like the sum of terms of a geometric sequence
a = r(1+i)^-1
common ratio = (1+i)^-1
number of terms = n
using the sum of n terms formula
sum = (r(1+i)^-1 (((1+i)^-1)^n - 1)/(
take out a common factor of r(1+i)^-n
we get
r(1+i)^-n [ (1+i)^(n-1) + (1+i)^(n-2) + .. + (1+i)^2 + (1+i)^1 + 1]
look at the terms in the square brackets and write them in reverse order
[1 + (1+i) + (1+i)^2 + ... + (1+i)^(n-1) }
there are n terms
with a = 1
and r = 1+i
sum(n) = a(r^n - 1)/(1+i - 1)
= 1( (1+i)^n - 1)/i
multiply this by our common factor:
( r(1+i)^-n )( (1+i)^n - 1)/i
= r( (1+i)^0 - (1+i)^-n)/i
= r( 1 - (1+i)^-n )/i
I think you forgot the "r" in your first sum
oh yea i did forget the r in my first sum thank you reiny i think i get it now
I started along one way, but changed my mind, part of the above was supposed to be deleted.
here is the final version:
r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + r(1+i)^-n
looks like the sum of terms of a geometric sequence
take out a common factor of r(1+i)^-n
we get
r(1+i)^-n [ (1+i)^(n-1) + (1+i)^(n-2) + .. + (1+i)^2 + (1+i)^1 + 1]
look at the terms in the square brackets and write them in reverse order
[1 + (1+i) + (1+i)^2 + ... + (1+i)^(n-1) }
there are n terms
with a = 1
and r = 1+i
sum(n) = a(r^n - 1)/(1+i - 1)
= 1( (1+i)^n - 1)/i
multiply this by our common factor:
( r(1+i)^-n )( (1+i)^n - 1)/i
= r( (1+i)^0 - (1+i)^-n)/i
= r( 1 - (1+i)^-n )/i
I think you forgot the "r" in your first sum
r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + r(1+i)^-n
r(1+i)^-n [ (1+i)^(n-1) + (1+i)^(n-2) + .. + (1+i)^2 + (1+i)^1 + 1]
how did you get the (1+i)^2? and why isnt there a (1+i)^ n+3?
yes, the line
r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + r(1+i)^-n
contains a typo
should have been
r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^-3 +...+ r(1+i)^-(n-1) + r(1+i)^-n
You should have realized that
if you take
r(1+i)^-n [ (1+i)^(n-1) + (1+i)^(n-2) + .. + (1+i)^2 + (1+i)^1 + 1]
and expand it, you will see why it is correct.
(remember, to multiply powers with the same base, we keep the base and add the exponents)