Posted by visoth on .
1) equation; sum= (1(1+i)^n / i
2) equation; sum= r(1+i)^1 + r(1+i) ^2 + r(1+i)^3 +...+ r(1+i)^(n1) + r(1+i)^n
im suppose to simplify second equation into the first one but im messing up somewhere if anyone can tell me the first 34 steps i think i can get it from there thank you

calculus 
Steve,
(1u)^n / (1u) = 1+u+u^2+...+u^n1
That help? 
calculus 
visoth,
im assuming you replaced (1+i) with u right if so i thinki kinda get

calculus 
Reiny,
r(1+i)^1 + r(1+i) ^2 + r(1+i)^3 +...+ r(1+i)^(n1) + 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)^(n1) + (1+i)^(n2) + .. + (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)^(n1) }
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 
calculus 
visoth,
oh yea i did forget the r in my first sum thank you reiny i think i get it now

calculus 
Reiny,
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)^(n1) + 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)^(n1) + (1+i)^(n2) + .. + (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)^(n1) }
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 
calculus 
visoth,
r(1+i)^1 + r(1+i) ^2 + r(1+i)^3 +...+ r(1+i)^(n1) + r(1+i)^n
r(1+i)^n [ (1+i)^(n1) + (1+i)^(n2) + .. + (1+i)^2 + (1+i)^1 + 1]
how did you get the (1+i)^2? and why isnt there a (1+i)^ n+3? 
calculus 
Reiny,
yes, the line
r(1+i)^1 + r(1+i) ^2 + r(1+i)^3 +...+ r(1+i)^(n1) + 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)^(n1) + r(1+i)^n
You should have realized that
if you take
r(1+i)^n [ (1+i)^(n1) + (1+i)^(n2) + .. + (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)