Mathematics of Finance












0












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If you invest a dollar at “6% interest compounded monthly,” it amounts to $1.005^n$
dollars after $n$ months. If you invest 10 dollars at the beginning of each month for 10 years (120 months), how much will you have at the end of the 10 years? I know the answer is $1646.99 but do not know how to get this answer using the compound interest formula. Please explain.






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  • $begingroup$
    We use dollar signs to set off MathJax, which is why your post changes from normal to italics. Please escape the dollar signs or escape them with backslashes.
    $endgroup$
    – Ross Millikan
    Jan 2 at 21:20










  • $begingroup$
    What have you tried? The compound interest formula is all you really need.
    $endgroup$
    – Neeyanth Kopparapu
    Jan 2 at 21:31










  • $begingroup$
    This is called Future value of annuity due
    $endgroup$
    – farruhota
    Jan 3 at 5:16
















0












$begingroup$


If you invest a dollar at “6% interest compounded monthly,” it amounts to $1.005^n$
dollars after $n$ months. If you invest 10 dollars at the beginning of each month for 10 years (120 months), how much will you have at the end of the 10 years? I know the answer is $1646.99 but do not know how to get this answer using the compound interest formula. Please explain.






finance







share|cite|improve this question











$endgroup$












  • $begingroup$
    We use dollar signs to set off MathJax, which is why your post changes from normal to italics. Please escape the dollar signs or escape them with backslashes.
    $endgroup$
    – Ross Millikan
    Jan 2 at 21:20










  • $begingroup$
    What have you tried? The compound interest formula is all you really need.
    $endgroup$
    – Neeyanth Kopparapu
    Jan 2 at 21:31










  • $begingroup$
    This is called Future value of annuity due
    $endgroup$
    – farruhota
    Jan 3 at 5:16














0












0








0





$begingroup$


If you invest a dollar at “6% interest compounded monthly,” it amounts to $1.005^n$
dollars after $n$ months. If you invest 10 dollars at the beginning of each month for 10 years (120 months), how much will you have at the end of the 10 years? I know the answer is $1646.99 but do not know how to get this answer using the compound interest formula. Please explain.






finance







share|cite|improve this question











$endgroup$




If you invest a dollar at “6% interest compounded monthly,” it amounts to $1.005^n$
dollars after $n$ months. If you invest 10 dollars at the beginning of each month for 10 years (120 months), how much will you have at the end of the 10 years? I know the answer is $1646.99 but do not know how to get this answer using the compound interest formula. Please explain.






finance




finance






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edited Jan 2 at 21:51









Madarb

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asked Jan 2 at 21:18









Eric BrownEric Brown

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  • $begingroup$
    We use dollar signs to set off MathJax, which is why your post changes from normal to italics. Please escape the dollar signs or escape them with backslashes.
    $endgroup$
    – Ross Millikan
    Jan 2 at 21:20










  • $begingroup$
    What have you tried? The compound interest formula is all you really need.
    $endgroup$
    – Neeyanth Kopparapu
    Jan 2 at 21:31










  • $begingroup$
    This is called Future value of annuity due
    $endgroup$
    – farruhota
    Jan 3 at 5:16


















  • $begingroup$
    We use dollar signs to set off MathJax, which is why your post changes from normal to italics. Please escape the dollar signs or escape them with backslashes.
    $endgroup$
    – Ross Millikan
    Jan 2 at 21:20










  • $begingroup$
    What have you tried? The compound interest formula is all you really need.
    $endgroup$
    – Neeyanth Kopparapu
    Jan 2 at 21:31










  • $begingroup$
    This is called Future value of annuity due
    $endgroup$
    – farruhota
    Jan 3 at 5:16
















$begingroup$
We use dollar signs to set off MathJax, which is why your post changes from normal to italics. Please escape the dollar signs or escape them with backslashes.
$endgroup$
– Ross Millikan
Jan 2 at 21:20




$begingroup$
We use dollar signs to set off MathJax, which is why your post changes from normal to italics. Please escape the dollar signs or escape them with backslashes.
$endgroup$
– Ross Millikan
Jan 2 at 21:20












$begingroup$
What have you tried? The compound interest formula is all you really need.
$endgroup$
– Neeyanth Kopparapu
Jan 2 at 21:31




$begingroup$
What have you tried? The compound interest formula is all you really need.
$endgroup$
– Neeyanth Kopparapu
Jan 2 at 21:31












$begingroup$
This is called Future value of annuity due
$endgroup$
– farruhota
Jan 3 at 5:16




$begingroup$
This is called Future value of annuity due
$endgroup$
– farruhota
Jan 3 at 5:16










1 Answer
1






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2












$begingroup$

Your first deposit compounds for 120 months.
The second for 119 months.
Your last deposit for 1 month.



The future value of 120 deposits.



$10(1.005)^{120} + 10(1.005)^{119} + cdots + 10(1.005)$



or



$sum_{n=1}^{120} 10(1.005)^n$



This is the sum of a geometric progression.



$sum_{n=1}^{m} y^n = frac {y(y^m-1)}{y-1}$



to find this formula multiply by $frac {1-y}{1-y}$



$frac {1}{1-y}(1-y)(y+y^2 + y^3 + cdots + y^m) = frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + cdots - y^{m+1})$



The expression "telescopes" leaving:



$frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.



plugging the numbers for this problem.



$10frac {1.005(1.005^{120} - 1)}{0.005}$






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    1 Answer
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    1 Answer
    1






    active

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    active

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    2












    $begingroup$

    Your first deposit compounds for 120 months.
    The second for 119 months.
    Your last deposit for 1 month.



    The future value of 120 deposits.



    $10(1.005)^{120} + 10(1.005)^{119} + cdots + 10(1.005)$



    or



    $sum_{n=1}^{120} 10(1.005)^n$



    This is the sum of a geometric progression.



    $sum_{n=1}^{m} y^n = frac {y(y^m-1)}{y-1}$



    to find this formula multiply by $frac {1-y}{1-y}$



    $frac {1}{1-y}(1-y)(y+y^2 + y^3 + cdots + y^m) = frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + cdots - y^{m+1})$



    The expression "telescopes" leaving:



    $frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.



    plugging the numbers for this problem.



    $10frac {1.005(1.005^{120} - 1)}{0.005}$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Your first deposit compounds for 120 months.
      The second for 119 months.
      Your last deposit for 1 month.



      The future value of 120 deposits.



      $10(1.005)^{120} + 10(1.005)^{119} + cdots + 10(1.005)$



      or



      $sum_{n=1}^{120} 10(1.005)^n$



      This is the sum of a geometric progression.



      $sum_{n=1}^{m} y^n = frac {y(y^m-1)}{y-1}$



      to find this formula multiply by $frac {1-y}{1-y}$



      $frac {1}{1-y}(1-y)(y+y^2 + y^3 + cdots + y^m) = frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + cdots - y^{m+1})$



      The expression "telescopes" leaving:



      $frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.



      plugging the numbers for this problem.



      $10frac {1.005(1.005^{120} - 1)}{0.005}$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Your first deposit compounds for 120 months.
        The second for 119 months.
        Your last deposit for 1 month.



        The future value of 120 deposits.



        $10(1.005)^{120} + 10(1.005)^{119} + cdots + 10(1.005)$



        or



        $sum_{n=1}^{120} 10(1.005)^n$



        This is the sum of a geometric progression.



        $sum_{n=1}^{m} y^n = frac {y(y^m-1)}{y-1}$



        to find this formula multiply by $frac {1-y}{1-y}$



        $frac {1}{1-y}(1-y)(y+y^2 + y^3 + cdots + y^m) = frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + cdots - y^{m+1})$



        The expression "telescopes" leaving:



        $frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.



        plugging the numbers for this problem.



        $10frac {1.005(1.005^{120} - 1)}{0.005}$






        share|cite|improve this answer











        $endgroup$



        Your first deposit compounds for 120 months.
        The second for 119 months.
        Your last deposit for 1 month.



        The future value of 120 deposits.



        $10(1.005)^{120} + 10(1.005)^{119} + cdots + 10(1.005)$



        or



        $sum_{n=1}^{120} 10(1.005)^n$



        This is the sum of a geometric progression.



        $sum_{n=1}^{m} y^n = frac {y(y^m-1)}{y-1}$



        to find this formula multiply by $frac {1-y}{1-y}$



        $frac {1}{1-y}(1-y)(y+y^2 + y^3 + cdots + y^m) = frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + cdots - y^{m+1})$



        The expression "telescopes" leaving:



        $frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.



        plugging the numbers for this problem.



        $10frac {1.005(1.005^{120} - 1)}{0.005}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 22:37

























        answered Jan 2 at 22:03









        Doug MDoug M

        44.5k31854




        44.5k31854






























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