calculus

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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)^-(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 -

    (1-u)^n / (1-u) = 1+u+u^2+...+u^n-1

    That help?

  • calculus -

    im assuming you replaced (1+i) with u right if so i thinki kinda get

  • calculus -

    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

  • calculus -

    oh yea i did forget the r in my first sum thank you reiny i think i get it now

  • calculus -

    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

  • calculus -

    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?

  • calculus -

    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)

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