Why is this equality wrong? Can't find a proof












1












$begingroup$


I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$



Edit: They are equal, thanks for the help






linear-algebra summation proof-explanation equivalence-relations







share|cite|improve this question









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  • 3




    $begingroup$
    Prove by induction on $xgeq 1$.
    $endgroup$
    – Wuestenfux
    1 hour ago










  • $begingroup$
    Did you really mean to say "(...) are not equal, I am certain they are"?
    $endgroup$
    – StackTD
    1 hour ago










  • $begingroup$
    Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
    $endgroup$
    – Ned
    1 hour ago










  • $begingroup$
    If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
    $endgroup$
    – KM101
    1 hour ago


















1












$begingroup$


I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$



Edit: They are equal, thanks for the help






linear-algebra summation proof-explanation equivalence-relations







share|cite|improve this question









New contributor




Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    Prove by induction on $xgeq 1$.
    $endgroup$
    – Wuestenfux
    1 hour ago










  • $begingroup$
    Did you really mean to say "(...) are not equal, I am certain they are"?
    $endgroup$
    – StackTD
    1 hour ago










  • $begingroup$
    Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
    $endgroup$
    – Ned
    1 hour ago










  • $begingroup$
    If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
    $endgroup$
    – KM101
    1 hour ago
















1












1








1





$begingroup$


I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$



Edit: They are equal, thanks for the help






linear-algebra summation proof-explanation equivalence-relations







share|cite|improve this question









New contributor




Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$



Edit: They are equal, thanks for the help






linear-algebra summation proof-explanation equivalence-relations




linear-algebra summation proof-explanation equivalence-relations






share|cite|improve this question









New contributor




Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







Jack Möller













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asked 1 hour ago









Jack MöllerJack Möller

62




62




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Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    $begingroup$
    Prove by induction on $xgeq 1$.
    $endgroup$
    – Wuestenfux
    1 hour ago










  • $begingroup$
    Did you really mean to say "(...) are not equal, I am certain they are"?
    $endgroup$
    – StackTD
    1 hour ago










  • $begingroup$
    Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
    $endgroup$
    – Ned
    1 hour ago










  • $begingroup$
    If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
    $endgroup$
    – KM101
    1 hour ago
















  • 3




    $begingroup$
    Prove by induction on $xgeq 1$.
    $endgroup$
    – Wuestenfux
    1 hour ago










  • $begingroup$
    Did you really mean to say "(...) are not equal, I am certain they are"?
    $endgroup$
    – StackTD
    1 hour ago










  • $begingroup$
    Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
    $endgroup$
    – Ned
    1 hour ago










  • $begingroup$
    If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
    $endgroup$
    – KM101
    1 hour ago










3




3




$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
1 hour ago




$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
1 hour ago












$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
1 hour ago




$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
1 hour ago












$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
1 hour ago




$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
1 hour ago












$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago






$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago












5 Answers
5






active

oldest

votes


















4












$begingroup$

$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    The easiest way is to notice that
    $$
    sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
    $$
    which is a telescoping series.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Use induction to prove it:




      • Show it is true for $x=1$


      • Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$


      • Then show that it is true for $n=k+1$ by using the above formula.







      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        You can prove this equality with the following method:
        x belongs to N*
        For x=1, 2^0 = 2^1 - 1 = 1
        For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
        .
        .
        .
        Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),



        Proof



        Thus, the equality holds for all x in N*






        share|cite|improve this answer








        New contributor




        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$





















          0












          $begingroup$


          Applying the finite geometric series formula we obtain
          begin{align*}
          sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
          end{align*}







          share|cite|improve this answer









          $endgroup$













            Your Answer





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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            $$
            S(x) = sum_{i=0}^{x-1} 2^{i} \
            2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
            $$
            so this is your proof that indeed $ S(x) = 2^x - 1 $.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              $$
              S(x) = sum_{i=0}^{x-1} 2^{i} \
              2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
              $$
              so this is your proof that indeed $ S(x) = 2^x - 1 $.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                $$
                S(x) = sum_{i=0}^{x-1} 2^{i} \
                2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
                $$
                so this is your proof that indeed $ S(x) = 2^x - 1 $.






                share|cite|improve this answer











                $endgroup$



                $$
                S(x) = sum_{i=0}^{x-1} 2^{i} \
                2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
                $$
                so this is your proof that indeed $ S(x) = 2^x - 1 $.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 58 mins ago

























                answered 1 hour ago









                AndreasAndreas

                7,8581037




                7,8581037























                    3












                    $begingroup$

                    The easiest way is to notice that
                    $$
                    sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
                    $$
                    which is a telescoping series.






                    share|cite|improve this answer









                    $endgroup$


















                      3












                      $begingroup$

                      The easiest way is to notice that
                      $$
                      sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
                      $$
                      which is a telescoping series.






                      share|cite|improve this answer









                      $endgroup$
















                        3












                        3








                        3





                        $begingroup$

                        The easiest way is to notice that
                        $$
                        sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
                        $$
                        which is a telescoping series.






                        share|cite|improve this answer









                        $endgroup$



                        The easiest way is to notice that
                        $$
                        sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
                        $$
                        which is a telescoping series.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 1 hour ago









                        Foobaz JohnFoobaz John

                        21.7k41352




                        21.7k41352























                            0












                            $begingroup$

                            Use induction to prove it:




                            • Show it is true for $x=1$


                            • Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$


                            • Then show that it is true for $n=k+1$ by using the above formula.







                            share|cite|improve this answer









                            $endgroup$


















                              0












                              $begingroup$

                              Use induction to prove it:




                              • Show it is true for $x=1$


                              • Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$


                              • Then show that it is true for $n=k+1$ by using the above formula.







                              share|cite|improve this answer









                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                Use induction to prove it:




                                • Show it is true for $x=1$


                                • Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$


                                • Then show that it is true for $n=k+1$ by using the above formula.







                                share|cite|improve this answer









                                $endgroup$



                                Use induction to prove it:




                                • Show it is true for $x=1$


                                • Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$


                                • Then show that it is true for $n=k+1$ by using the above formula.








                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 1 hour ago









                                Ben CrossleyBen Crossley

                                787318




                                787318























                                    0












                                    $begingroup$

                                    You can prove this equality with the following method:
                                    x belongs to N*
                                    For x=1, 2^0 = 2^1 - 1 = 1
                                    For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
                                    .
                                    .
                                    .
                                    Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),



                                    Proof



                                    Thus, the equality holds for all x in N*






                                    share|cite|improve this answer








                                    New contributor




                                    Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$


















                                      0












                                      $begingroup$

                                      You can prove this equality with the following method:
                                      x belongs to N*
                                      For x=1, 2^0 = 2^1 - 1 = 1
                                      For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
                                      .
                                      .
                                      .
                                      Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),



                                      Proof



                                      Thus, the equality holds for all x in N*






                                      share|cite|improve this answer








                                      New contributor




                                      Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        You can prove this equality with the following method:
                                        x belongs to N*
                                        For x=1, 2^0 = 2^1 - 1 = 1
                                        For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
                                        .
                                        .
                                        .
                                        Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),



                                        Proof



                                        Thus, the equality holds for all x in N*






                                        share|cite|improve this answer








                                        New contributor




                                        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        $endgroup$



                                        You can prove this equality with the following method:
                                        x belongs to N*
                                        For x=1, 2^0 = 2^1 - 1 = 1
                                        For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
                                        .
                                        .
                                        .
                                        Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),



                                        Proof



                                        Thus, the equality holds for all x in N*







                                        share|cite|improve this answer








                                        New contributor




                                        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        share|cite|improve this answer



                                        share|cite|improve this answer






                                        New contributor




                                        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        answered 1 hour ago









                                        Wane MamadouWane Mamadou

                                        1




                                        1




                                        New contributor




                                        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                        New contributor





                                        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.























                                            0












                                            $begingroup$


                                            Applying the finite geometric series formula we obtain
                                            begin{align*}
                                            sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
                                            end{align*}







                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$


                                              Applying the finite geometric series formula we obtain
                                              begin{align*}
                                              sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
                                              end{align*}







                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$


                                                Applying the finite geometric series formula we obtain
                                                begin{align*}
                                                sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
                                                end{align*}







                                                share|cite|improve this answer









                                                $endgroup$




                                                Applying the finite geometric series formula we obtain
                                                begin{align*}
                                                sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
                                                end{align*}








                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 10 mins ago









                                                Markus ScheuerMarkus Scheuer

                                                60.6k455145




                                                60.6k455145






















                                                    Jack Möller is a new contributor. Be nice, and check out our Code of Conduct.










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                                                    Jack Möller is a new contributor. Be nice, and check out our Code of Conduct.
















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                                                    Gregoriusmis

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                                                    Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl...
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                                                    Goes

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                                                    Zie Goes (doorverwijspagina) voor andere betekenissen van Goes. Goes Plaats in Nederland (Details) (Details) Situering Provincie Zeeland Gemeente Goes Coördinaten 51° 30′ NB, 3° 53′ OL Algemeen Inwoners (01-01-2018) 37.636 [1] Overig Netnummer 0113 Belangrijke verkeersaders A58, N254, N256, N289, NS Zeeuwse Lijn Website www.goes.nl Portaal     Nederland Stadshaven van Goes Het stadhuis Grote of Maria Magdalenakerk De watertoren Westerkerk De Koornbloem De televisietoren Getijwaterkorenmolen 't Soepuus Trapgevels in Goes Stadsgracht ("de Veste") Kijk op de Nieuwstraat vanaf de Beestenmarkt Cultuurhuis in Amsterdamse Schoolstijl Voormalige RK ULO Goes (   uitspraak   (info / uitleg) ) (Zeeuws: Hoes ) is een stad op het schiereiland Zuid-Beveland, centraal gelegen in de Nederlandse provincie Zeeland. De s...
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