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







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












$begingroup$


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

        68.7k22685




        68.7k22685























            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

            8,05121738




            8,05121738












            • $begingroup$
              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



            -

            Popular posts from this blog

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more