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Math Help - simplification

  1. #1
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    simplification

    How can i count this easier???

    2500*(1,0525)^5+2500*(1,0525)^4+2500*(1.0525)^3+25 00*(1.0525)^2+2500*(1,0525)^1= 14612.10382

    How could i count this without writing allt this down?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rambo View Post
    How can i count this easier???

    2500*(1,0525)^5+2500*(1,0525)^4+2500*(1.0525)^3+25 00*(1.0525)^2+2500*(1,0525)^1= 14612.10382

    How could i count this without writing allt this down?
    The only way I can think of to help with this is to factor a bit:
    (2500 \cdot 10525)(10525^4 + 10525^3 + 10525^2 + 10525 + 1)

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    The only way I can think of to help with this is to factor a bit:
    (2500 \cdot 10525)(10525^4 + 10525^3 + 10525^2 + 10525 + 1)

    -Dan
    thats not correct that gets a other answer 17679.91359
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rambo View Post
    thats not correct that gets a other answer 17679.91359
    Funny because I'm getting 322914659677371145062500 doing it both ways. Not what you got. How did you get such a small answer?

    -Dan

    Edit: Oh, I get it now. It isn't 10125 it is 1.0125. (The comma-decimal point thing gets me in the end again!) Still, I get the same answer using both expressions: 14612.103817994

    -Dan
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  5. #5
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    Rambo,

    That is the Accumulated Value of an annuity due series. There is a standard formula for that:

    Your Common Term or payment is 2500.

    Your Sum of the Series:

    Payment * ((1 + i)^n - 1) * (1 + i)/i

    In your textbook, they might refer to d, which is i/(1 + i).

    That being said, i is .0525, payment is 2500, and n is 5. Using that formula, I match your numbers to the decimal place.
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  6. #6
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    Quote Originally Posted by Rambo View Post
    How can i count this easier???

    2500*(1,0525)^5+2500*(1,0525)^4+2500*(1.0525)^3+25 00*(1.0525)^2+2500*(1,0525)^1= 14612.10382

    How could i count this without writing allt this down?
    First I assume that you use the comma as a decimal point.

    Second there must be a typo because the LHS yields 14612.10381

    Third: you can factor out 2500:

    2500*(1.0525)^5+2500*(1.0525)^4+2500*(1.0525)^3+ 2500*(1.0525)^2+2500*(1.0525)^1 =  2500*(1.0525)^5+(1.0525)^4+(1.0525)^3+ (1.0525)^2+(1.0525)^1)= 14612.10381
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  7. #7
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    if i use your formula

    2500*((1+0,525)^5*(1+0,525)/0,525= 59896.26856



    The answer will be wrong!!!!
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  8. #8
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    Quote Originally Posted by Rambo View Post
    2500*((1+0,525)^5*(1+0,525)/0,525= 59896.26856



    The answer will be wrong!!!!
    Did you read it right? Payment * (1 + i) * ((1 + i)^n - 1)/i

    2500 * 1.0525 * ((1.0525)^5 - 1)/.0525

    This is 14612.10

    It does what Earthbot's series does. You mentioned you like using series better, so I'd go with Earthbot's suggestion.
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  9. #9
    Super Member wingless's Avatar
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    2500(1,0525)^5+2500(1,0525)^4+2500(1.0525)^3+2500(  1.0525)^2+2500(1,0525)^1

    2500 ({1,0525}^1+{1.0525}^2+{1.0525}^3+{1,0525}^4+{1,05  25}^5)

    2500 \sum_{k=1}^{5} 1.0525^k

    \sum_{k=1}^{n} r^k = \frac{r^{n+1}-1}{r-1}-1

    2500 \sum_{k=1}^{5} 1.0525^k = 2500 \left (\frac{{1.0525}^{6}-1}{{1.0525}-1}-1 \right )

    2500 \left ( \frac{0.35935}{0.0525}-1 \right ) = 14612.1
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  10. #10
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    Hello, Rambo!

    It helps if you know this identity:

    . . \frac{X^n - 1}{X - 1} \;=\;X^{n-1} + X^{n-2} + X^{n-3} + \hdots + X^2 + X+ 1


    Evaluate:
    2500(1.0525)^5+2500(1.0525)^4+2500(1.0525)^3+2500(  1.0525)^2+2500(1,0525)^1

    Factor: . 2500(1.0525)\,\bigg[1.0525^4 + 1.0525^3 + 1.0525^2 + 1.0525 + 1\bigg]

    . . . . = \;2500(1.0525)\left[\frac{1.0525^5 - 1}{1.0525 - 1}\right]

    . . . . = \;(2631.25)\,\frac{0.291547915}{0.0525}

    . . . . = \;14612.10382

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