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July 7, 2015

July 7, 2015

Posted by **visoth** on Thursday, January 31, 2013 at 10:21am.

2) equation; sum= r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + 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 3-4 steps i think i can get it from there thank you

- calculus -
**Steve**, Thursday, January 31, 2013 at 10:37am(1-u)^n / (1-u) = 1+u+u^2+...+u^n-1

That help?

- calculus -
**visoth**, Thursday, January 31, 2013 at 10:40amim assuming you replaced (1+i) with u right if so i thinki kinda get

- calculus -
**Reiny**, Thursday, January 31, 2013 at 10:45amr(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

- calculus -
**visoth**, Thursday, January 31, 2013 at 10:47amoh yea i did forget the r in my first sum thank you reiny i think i get it now

- calculus -
**Reiny**, Thursday, January 31, 2013 at 10:47amI 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

- calculus -
**visoth**, Thursday, January 31, 2013 at 10:55amr(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?

- calculus -
**Reiny**, Thursday, January 31, 2013 at 11:09amyes, 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)