calculus

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

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

    That help?

    posted by Steve
  2. im assuming you replaced (1+i) with u right if so i thinki kinda get

    posted by visoth
  3. 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

    posted by Reiny
  4. oh yea i did forget the r in my first sum thank you reiny i think i get it now

    posted by visoth
  5. 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

    posted by Reiny
  6. 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?

    posted by visoth
  7. 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)

    posted by Reiny

Respond to this Question

First Name

Your Answer

Similar Questions

  1. Find the sum of thuis question

    Can someone find the sum for this equation please. (X^2)/(X^2-16)=(5X+4)/(X^2-16) I don't think it is finding the sum, but I can simplify the equation. (X^2)/(X^2-16)=(5X+4)/(X^2-16) First, multiply both sides by X^-16. X^2 = 5X+4
  2. Find the sum

    Can someone how to find the sum and simpliest form of this equation?(x+4)/5+(2x+1)/5 It is not an equation because there is no = sign. It is a expression. You can simplify it a bit by realising it has a common denominator in both
  3. ALG 2 PLZ PLZ HELP

    1.)The product of 14 and n is -28. Write the algebraic equation. 2.)Four times a certain number is the same as the number increased by 78 3.)A blue bike is $14 less than a red bike. The sum of their prices is $300. How much is the
  4. Calculus again

    Find the equation of the tangent to y=2-xe^x at the point where x=0 So I found the first derivative by doing the sum rule and product rule. I got e^(x)(x-1) Then I found the slope by substituting 0 for x and got -1. Then I found y
  5. Math; Please help and thank you

    sin(x+(pi/4))- sin(x-(pi/4))=1 1. Apply the sum and difference formulas for sine to simplify the left-hand side of the equation. 2. find all the solutions you found to the equation you found.
  6. Calculus

    Find the sum of two numbers whose sum is 70 and whose product is as large as possible. Ok i got that one equation is y = 70 - x...but how do I show an equation for a product which is as large as possible. The product is P = x (70
  7. algebra

    add equation (5x^8 -5z^4+5z^2+3)+(3z^6+4 z^4-9z) simplify answer to sum
  8. math urgent :(

    1.)A number decreased by 32 is -58. Find the number. The number is ____. (Give only the value of the number as your answer. 2.)The product of 14 and n is -28. Write the algebraic equation 3.)A blue bike is $14 less than a red
  9. math ALG 2!

    1.)A number decreased by 32 is -58. Find the number. The number is ____. (Give only the value of the number as your answer. 2.)The product of 14 and n is -28. Write the algebraic equation 3.)A blue bike is $14 less than a red
  10. math ALG 2!

    1.)A number decreased by 32 is -58. Find the number. The number is ____. (Give only the value of the number as your answer. 2.)The product of 14 and n is -28. Write the algebraic equation 3.)A blue bike is $14 less than a red

More Similar Questions