Easiest way to compute large set of numbers [closed]












0












$begingroup$


4−12+36−108+324−...+236,196



What is the common trick to compute numbers with the same path ?



Here the path is have x-3x+3(3x)-.....



Thanks!






discrete-mathematics







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closed as off-topic by Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138 Dec 7 '18 at 2:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    en.wikipedia.org/wiki/Geometric_series
    $endgroup$
    – vadim123
    Dec 6 '18 at 23:08










  • $begingroup$
    Do you know the trick that $frac {a^{n+1} - 1}{a-1} =1 + a + a^2 + ..... + a^n$?
    $endgroup$
    – fleablood
    Dec 6 '18 at 23:19
















0












$begingroup$


4−12+36−108+324−...+236,196



What is the common trick to compute numbers with the same path ?



Here the path is have x-3x+3(3x)-.....



Thanks!






discrete-mathematics







share|cite|improve this question









$endgroup$



closed as off-topic by Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138 Dec 7 '18 at 2:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    en.wikipedia.org/wiki/Geometric_series
    $endgroup$
    – vadim123
    Dec 6 '18 at 23:08










  • $begingroup$
    Do you know the trick that $frac {a^{n+1} - 1}{a-1} =1 + a + a^2 + ..... + a^n$?
    $endgroup$
    – fleablood
    Dec 6 '18 at 23:19














0












0








0





$begingroup$


4−12+36−108+324−...+236,196



What is the common trick to compute numbers with the same path ?



Here the path is have x-3x+3(3x)-.....



Thanks!






discrete-mathematics







share|cite|improve this question









$endgroup$




4−12+36−108+324−...+236,196



What is the common trick to compute numbers with the same path ?



Here the path is have x-3x+3(3x)-.....



Thanks!






discrete-mathematics




discrete-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 23:07









Tom1999Tom1999

445




445




closed as off-topic by Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138 Dec 7 '18 at 2:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138 Dec 7 '18 at 2:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, José Carlos Santos, T. Bongers, Leucippus, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    en.wikipedia.org/wiki/Geometric_series
    $endgroup$
    – vadim123
    Dec 6 '18 at 23:08










  • $begingroup$
    Do you know the trick that $frac {a^{n+1} - 1}{a-1} =1 + a + a^2 + ..... + a^n$?
    $endgroup$
    – fleablood
    Dec 6 '18 at 23:19














  • 2




    $begingroup$
    en.wikipedia.org/wiki/Geometric_series
    $endgroup$
    – vadim123
    Dec 6 '18 at 23:08










  • $begingroup$
    Do you know the trick that $frac {a^{n+1} - 1}{a-1} =1 + a + a^2 + ..... + a^n$?
    $endgroup$
    – fleablood
    Dec 6 '18 at 23:19








2




2




$begingroup$
en.wikipedia.org/wiki/Geometric_series
$endgroup$
– vadim123
Dec 6 '18 at 23:08




$begingroup$
en.wikipedia.org/wiki/Geometric_series
$endgroup$
– vadim123
Dec 6 '18 at 23:08












$begingroup$
Do you know the trick that $frac {a^{n+1} - 1}{a-1} =1 + a + a^2 + ..... + a^n$?
$endgroup$
– fleablood
Dec 6 '18 at 23:19




$begingroup$
Do you know the trick that $frac {a^{n+1} - 1}{a-1} =1 + a + a^2 + ..... + a^n$?
$endgroup$
– fleablood
Dec 6 '18 at 23:19










2 Answers
2






active

oldest

votes


















2












$begingroup$

So the number is



$4 - 3*4 + 3^2*4 - ...... + 3^10*4 = $



$4(1 - 3 + 3^2 - ...... + 3^{10})=$



$4(1 + (-3)^1 +(-3)^2 + ..... + (-3)^{10}) = $



$4frac {1+ (-3)^{11}}{1 - (-3)}=$ (do you know that trick? [1])



$4frac {1 - 3^{11}}{4} = 1 - 177147= -177146$



====



[1]$(1 + a + a^2 + ..... + a^n)(1-a)=$



$(1 + a + a^2 + ..... + a^n)-(a + a^2 + a^2 + ..... + a^{n+1})=$



$(1- a^{n+1})$ and therefore:



$ 1 + a + a^2 + ..... + a^n= frac {1 - a^{n+1}}{1-a} $



And $r + ra + ra^2 + ..... + ra^n = rfrac {1 - a^{n+1}}{1-a}$



It's a very handy trick. It called a geometric series






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    This is a Geometric Series, i.e. a sum of a sequence of numbers where each number is obtained from the previous by multiplying by the same number each time. For example, the sum $1-2+4-8+cdots$ is a Geometric Series since you get each number by multiplication by $-2$ whereas $1+2+4+8+24+72+cdots$ is not Geometric since there is not a consistent number to get to the next number each time.



    If you are summing a finite number of geometric terms, the sum is
    $$
    a+ar+ar^2+cdots= a left( dfrac{1-r^n}{1-r} right),
    $$
    where $a$ is the first number, $r$ is the number you multiply by, i.e. the common ratio, and $n$ is the number of terms. For example,
    $$
    1-3+9-27+81= 1 cdot left(dfrac{1-(-3)^5}{1-(-3)} right)=
    $$



    If you use an infinite amount of terms, then the sum is
    $$
    a+ar+ar^2+cdots+ar^3+cdots= dfrac{a}{1-r},
    $$
    assuming that the number $r$ you multiply by has the property that $-1<r<1$, otherwise the sum will not converge. For example,
    $$
    3- dfrac{3}{2} + dfrac{3}{4} - dfrac{3}{8} + cdots = dfrac{3}{1-(-1/2)}= dfrac{3}{3/2}= 3.
    $$
    For more on this, check out the Wiki page: Geometric Series






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thank you guys!!!
      $endgroup$
      – Tom1999
      Dec 6 '18 at 23:23


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    So the number is



    $4 - 3*4 + 3^2*4 - ...... + 3^10*4 = $



    $4(1 - 3 + 3^2 - ...... + 3^{10})=$



    $4(1 + (-3)^1 +(-3)^2 + ..... + (-3)^{10}) = $



    $4frac {1+ (-3)^{11}}{1 - (-3)}=$ (do you know that trick? [1])



    $4frac {1 - 3^{11}}{4} = 1 - 177147= -177146$



    ====



    [1]$(1 + a + a^2 + ..... + a^n)(1-a)=$



    $(1 + a + a^2 + ..... + a^n)-(a + a^2 + a^2 + ..... + a^{n+1})=$



    $(1- a^{n+1})$ and therefore:



    $ 1 + a + a^2 + ..... + a^n= frac {1 - a^{n+1}}{1-a} $



    And $r + ra + ra^2 + ..... + ra^n = rfrac {1 - a^{n+1}}{1-a}$



    It's a very handy trick. It called a geometric series






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      So the number is



      $4 - 3*4 + 3^2*4 - ...... + 3^10*4 = $



      $4(1 - 3 + 3^2 - ...... + 3^{10})=$



      $4(1 + (-3)^1 +(-3)^2 + ..... + (-3)^{10}) = $



      $4frac {1+ (-3)^{11}}{1 - (-3)}=$ (do you know that trick? [1])



      $4frac {1 - 3^{11}}{4} = 1 - 177147= -177146$



      ====



      [1]$(1 + a + a^2 + ..... + a^n)(1-a)=$



      $(1 + a + a^2 + ..... + a^n)-(a + a^2 + a^2 + ..... + a^{n+1})=$



      $(1- a^{n+1})$ and therefore:



      $ 1 + a + a^2 + ..... + a^n= frac {1 - a^{n+1}}{1-a} $



      And $r + ra + ra^2 + ..... + ra^n = rfrac {1 - a^{n+1}}{1-a}$



      It's a very handy trick. It called a geometric series






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        So the number is



        $4 - 3*4 + 3^2*4 - ...... + 3^10*4 = $



        $4(1 - 3 + 3^2 - ...... + 3^{10})=$



        $4(1 + (-3)^1 +(-3)^2 + ..... + (-3)^{10}) = $



        $4frac {1+ (-3)^{11}}{1 - (-3)}=$ (do you know that trick? [1])



        $4frac {1 - 3^{11}}{4} = 1 - 177147= -177146$



        ====



        [1]$(1 + a + a^2 + ..... + a^n)(1-a)=$



        $(1 + a + a^2 + ..... + a^n)-(a + a^2 + a^2 + ..... + a^{n+1})=$



        $(1- a^{n+1})$ and therefore:



        $ 1 + a + a^2 + ..... + a^n= frac {1 - a^{n+1}}{1-a} $



        And $r + ra + ra^2 + ..... + ra^n = rfrac {1 - a^{n+1}}{1-a}$



        It's a very handy trick. It called a geometric series






        share|cite|improve this answer









        $endgroup$



        So the number is



        $4 - 3*4 + 3^2*4 - ...... + 3^10*4 = $



        $4(1 - 3 + 3^2 - ...... + 3^{10})=$



        $4(1 + (-3)^1 +(-3)^2 + ..... + (-3)^{10}) = $



        $4frac {1+ (-3)^{11}}{1 - (-3)}=$ (do you know that trick? [1])



        $4frac {1 - 3^{11}}{4} = 1 - 177147= -177146$



        ====



        [1]$(1 + a + a^2 + ..... + a^n)(1-a)=$



        $(1 + a + a^2 + ..... + a^n)-(a + a^2 + a^2 + ..... + a^{n+1})=$



        $(1- a^{n+1})$ and therefore:



        $ 1 + a + a^2 + ..... + a^n= frac {1 - a^{n+1}}{1-a} $



        And $r + ra + ra^2 + ..... + ra^n = rfrac {1 - a^{n+1}}{1-a}$



        It's a very handy trick. It called a geometric series







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 23:17









        fleabloodfleablood

        72k22687




        72k22687























            1












            $begingroup$

            This is a Geometric Series, i.e. a sum of a sequence of numbers where each number is obtained from the previous by multiplying by the same number each time. For example, the sum $1-2+4-8+cdots$ is a Geometric Series since you get each number by multiplication by $-2$ whereas $1+2+4+8+24+72+cdots$ is not Geometric since there is not a consistent number to get to the next number each time.



            If you are summing a finite number of geometric terms, the sum is
            $$
            a+ar+ar^2+cdots= a left( dfrac{1-r^n}{1-r} right),
            $$
            where $a$ is the first number, $r$ is the number you multiply by, i.e. the common ratio, and $n$ is the number of terms. For example,
            $$
            1-3+9-27+81= 1 cdot left(dfrac{1-(-3)^5}{1-(-3)} right)=
            $$



            If you use an infinite amount of terms, then the sum is
            $$
            a+ar+ar^2+cdots+ar^3+cdots= dfrac{a}{1-r},
            $$
            assuming that the number $r$ you multiply by has the property that $-1<r<1$, otherwise the sum will not converge. For example,
            $$
            3- dfrac{3}{2} + dfrac{3}{4} - dfrac{3}{8} + cdots = dfrac{3}{1-(-1/2)}= dfrac{3}{3/2}= 3.
            $$
            For more on this, check out the Wiki page: Geometric Series






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank you guys!!!
              $endgroup$
              – Tom1999
              Dec 6 '18 at 23:23
















            1












            $begingroup$

            This is a Geometric Series, i.e. a sum of a sequence of numbers where each number is obtained from the previous by multiplying by the same number each time. For example, the sum $1-2+4-8+cdots$ is a Geometric Series since you get each number by multiplication by $-2$ whereas $1+2+4+8+24+72+cdots$ is not Geometric since there is not a consistent number to get to the next number each time.



            If you are summing a finite number of geometric terms, the sum is
            $$
            a+ar+ar^2+cdots= a left( dfrac{1-r^n}{1-r} right),
            $$
            where $a$ is the first number, $r$ is the number you multiply by, i.e. the common ratio, and $n$ is the number of terms. For example,
            $$
            1-3+9-27+81= 1 cdot left(dfrac{1-(-3)^5}{1-(-3)} right)=
            $$



            If you use an infinite amount of terms, then the sum is
            $$
            a+ar+ar^2+cdots+ar^3+cdots= dfrac{a}{1-r},
            $$
            assuming that the number $r$ you multiply by has the property that $-1<r<1$, otherwise the sum will not converge. For example,
            $$
            3- dfrac{3}{2} + dfrac{3}{4} - dfrac{3}{8} + cdots = dfrac{3}{1-(-1/2)}= dfrac{3}{3/2}= 3.
            $$
            For more on this, check out the Wiki page: Geometric Series






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank you guys!!!
              $endgroup$
              – Tom1999
              Dec 6 '18 at 23:23














            1












            1








            1





            $begingroup$

            This is a Geometric Series, i.e. a sum of a sequence of numbers where each number is obtained from the previous by multiplying by the same number each time. For example, the sum $1-2+4-8+cdots$ is a Geometric Series since you get each number by multiplication by $-2$ whereas $1+2+4+8+24+72+cdots$ is not Geometric since there is not a consistent number to get to the next number each time.



            If you are summing a finite number of geometric terms, the sum is
            $$
            a+ar+ar^2+cdots= a left( dfrac{1-r^n}{1-r} right),
            $$
            where $a$ is the first number, $r$ is the number you multiply by, i.e. the common ratio, and $n$ is the number of terms. For example,
            $$
            1-3+9-27+81= 1 cdot left(dfrac{1-(-3)^5}{1-(-3)} right)=
            $$



            If you use an infinite amount of terms, then the sum is
            $$
            a+ar+ar^2+cdots+ar^3+cdots= dfrac{a}{1-r},
            $$
            assuming that the number $r$ you multiply by has the property that $-1<r<1$, otherwise the sum will not converge. For example,
            $$
            3- dfrac{3}{2} + dfrac{3}{4} - dfrac{3}{8} + cdots = dfrac{3}{1-(-1/2)}= dfrac{3}{3/2}= 3.
            $$
            For more on this, check out the Wiki page: Geometric Series






            share|cite|improve this answer









            $endgroup$



            This is a Geometric Series, i.e. a sum of a sequence of numbers where each number is obtained from the previous by multiplying by the same number each time. For example, the sum $1-2+4-8+cdots$ is a Geometric Series since you get each number by multiplication by $-2$ whereas $1+2+4+8+24+72+cdots$ is not Geometric since there is not a consistent number to get to the next number each time.



            If you are summing a finite number of geometric terms, the sum is
            $$
            a+ar+ar^2+cdots= a left( dfrac{1-r^n}{1-r} right),
            $$
            where $a$ is the first number, $r$ is the number you multiply by, i.e. the common ratio, and $n$ is the number of terms. For example,
            $$
            1-3+9-27+81= 1 cdot left(dfrac{1-(-3)^5}{1-(-3)} right)=
            $$



            If you use an infinite amount of terms, then the sum is
            $$
            a+ar+ar^2+cdots+ar^3+cdots= dfrac{a}{1-r},
            $$
            assuming that the number $r$ you multiply by has the property that $-1<r<1$, otherwise the sum will not converge. For example,
            $$
            3- dfrac{3}{2} + dfrac{3}{4} - dfrac{3}{8} + cdots = dfrac{3}{1-(-1/2)}= dfrac{3}{3/2}= 3.
            $$
            For more on this, check out the Wiki page: Geometric Series







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 23:18









            mathematics2x2lifemathematics2x2life

            8,06621739




            8,06621739












            • $begingroup$
              thank you guys!!!
              $endgroup$
              – Tom1999
              Dec 6 '18 at 23:23


















            • $begingroup$
              thank you guys!!!
              $endgroup$
              – Tom1999
              Dec 6 '18 at 23:23
















            $begingroup$
            thank you guys!!!
            $endgroup$
            – Tom1999
            Dec 6 '18 at 23:23




            $begingroup$
            thank you guys!!!
            $endgroup$
            – Tom1999
            Dec 6 '18 at 23:23



            -

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