Proving the geometric series $sum_{i=0}^n r^i = frac{1-r^{n+1}}{1-r}$ by induction for $ngeq 1$












1












$begingroup$


Let $P(n)$ be the statement
$$
P(n) : sumlimits^{n}_{i=0}r^i = dfrac{1-r^{1+n}}{1-r}text{ for all }n in mathbb{N}text{.}
$$
I am stuck at the base case:
$$P(1):1 + r = dfrac{1-r^2}{1-r}text{.}$$



I am stuck as to how I can show $P(1)$ is true.






proof-writing induction







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: $a^2-b^2=(a+b)(a-b)$
    $endgroup$
    – user105475
    Aug 14 '14 at 1:42










  • $begingroup$
    Base Case is P(0) not P(1)
    $endgroup$
    – user137481
    Aug 14 '14 at 1:42










  • $begingroup$
    @user137481 Natural Numbers start at 1 ... so should the base case not be P(1)
    $endgroup$
    – Guest
    Aug 14 '14 at 1:43






  • 1




    $begingroup$
    That depends on your teacher. In some courses, Natural Numbers start at 0
    $endgroup$
    – user137481
    Aug 14 '14 at 1:47






  • 1




    $begingroup$
    Note that the problem really should have specified that $rne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$.
    $endgroup$
    – André Nicolas
    Aug 14 '14 at 1:48
















1












$begingroup$


Let $P(n)$ be the statement
$$
P(n) : sumlimits^{n}_{i=0}r^i = dfrac{1-r^{1+n}}{1-r}text{ for all }n in mathbb{N}text{.}
$$
I am stuck at the base case:
$$P(1):1 + r = dfrac{1-r^2}{1-r}text{.}$$



I am stuck as to how I can show $P(1)$ is true.






proof-writing induction







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: $a^2-b^2=(a+b)(a-b)$
    $endgroup$
    – user105475
    Aug 14 '14 at 1:42










  • $begingroup$
    Base Case is P(0) not P(1)
    $endgroup$
    – user137481
    Aug 14 '14 at 1:42










  • $begingroup$
    @user137481 Natural Numbers start at 1 ... so should the base case not be P(1)
    $endgroup$
    – Guest
    Aug 14 '14 at 1:43






  • 1




    $begingroup$
    That depends on your teacher. In some courses, Natural Numbers start at 0
    $endgroup$
    – user137481
    Aug 14 '14 at 1:47






  • 1




    $begingroup$
    Note that the problem really should have specified that $rne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$.
    $endgroup$
    – André Nicolas
    Aug 14 '14 at 1:48














1












1








1


2



$begingroup$


Let $P(n)$ be the statement
$$
P(n) : sumlimits^{n}_{i=0}r^i = dfrac{1-r^{1+n}}{1-r}text{ for all }n in mathbb{N}text{.}
$$
I am stuck at the base case:
$$P(1):1 + r = dfrac{1-r^2}{1-r}text{.}$$



I am stuck as to how I can show $P(1)$ is true.






proof-writing induction







share|cite|improve this question











$endgroup$




Let $P(n)$ be the statement
$$
P(n) : sumlimits^{n}_{i=0}r^i = dfrac{1-r^{1+n}}{1-r}text{ for all }n in mathbb{N}text{.}
$$
I am stuck at the base case:
$$P(1):1 + r = dfrac{1-r^2}{1-r}text{.}$$



I am stuck as to how I can show $P(1)$ is true.






proof-writing induction




proof-writing induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 '15 at 16:36









Daniel W. Farlow

17.6k114488




17.6k114488










asked Aug 14 '14 at 1:40









GuestGuest

17313




17313








  • 3




    $begingroup$
    Hint: $a^2-b^2=(a+b)(a-b)$
    $endgroup$
    – user105475
    Aug 14 '14 at 1:42










  • $begingroup$
    Base Case is P(0) not P(1)
    $endgroup$
    – user137481
    Aug 14 '14 at 1:42










  • $begingroup$
    @user137481 Natural Numbers start at 1 ... so should the base case not be P(1)
    $endgroup$
    – Guest
    Aug 14 '14 at 1:43






  • 1




    $begingroup$
    That depends on your teacher. In some courses, Natural Numbers start at 0
    $endgroup$
    – user137481
    Aug 14 '14 at 1:47






  • 1




    $begingroup$
    Note that the problem really should have specified that $rne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$.
    $endgroup$
    – André Nicolas
    Aug 14 '14 at 1:48














  • 3




    $begingroup$
    Hint: $a^2-b^2=(a+b)(a-b)$
    $endgroup$
    – user105475
    Aug 14 '14 at 1:42










  • $begingroup$
    Base Case is P(0) not P(1)
    $endgroup$
    – user137481
    Aug 14 '14 at 1:42










  • $begingroup$
    @user137481 Natural Numbers start at 1 ... so should the base case not be P(1)
    $endgroup$
    – Guest
    Aug 14 '14 at 1:43






  • 1




    $begingroup$
    That depends on your teacher. In some courses, Natural Numbers start at 0
    $endgroup$
    – user137481
    Aug 14 '14 at 1:47






  • 1




    $begingroup$
    Note that the problem really should have specified that $rne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$.
    $endgroup$
    – André Nicolas
    Aug 14 '14 at 1:48








3




3




$begingroup$
Hint: $a^2-b^2=(a+b)(a-b)$
$endgroup$
– user105475
Aug 14 '14 at 1:42




$begingroup$
Hint: $a^2-b^2=(a+b)(a-b)$
$endgroup$
– user105475
Aug 14 '14 at 1:42












$begingroup$
Base Case is P(0) not P(1)
$endgroup$
– user137481
Aug 14 '14 at 1:42




$begingroup$
Base Case is P(0) not P(1)
$endgroup$
– user137481
Aug 14 '14 at 1:42












$begingroup$
@user137481 Natural Numbers start at 1 ... so should the base case not be P(1)
$endgroup$
– Guest
Aug 14 '14 at 1:43




$begingroup$
@user137481 Natural Numbers start at 1 ... so should the base case not be P(1)
$endgroup$
– Guest
Aug 14 '14 at 1:43




1




1




$begingroup$
That depends on your teacher. In some courses, Natural Numbers start at 0
$endgroup$
– user137481
Aug 14 '14 at 1:47




$begingroup$
That depends on your teacher. In some courses, Natural Numbers start at 0
$endgroup$
– user137481
Aug 14 '14 at 1:47




1




1




$begingroup$
Note that the problem really should have specified that $rne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$.
$endgroup$
– André Nicolas
Aug 14 '14 at 1:48




$begingroup$
Note that the problem really should have specified that $rne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$.
$endgroup$
– André Nicolas
Aug 14 '14 at 1:48










2 Answers
2






active

oldest

votes


















0












$begingroup$

Note: First note that you are really summing the geometric series ($a,rinmathbb{R}, aneq 0, rneq 1$)
$$
a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}=afrac{1-r^{n+1}}{1-r}
$$
with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.



Claim: Let $a$ and $r$ be real numbers with $aneq 0$ and $rneq 1$. Prove that for each integer $ngeq 1$ that
$$
a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}.
$$
Proof. Fix $rinmathbb{R}, rneq 1$, and let $S(n)$ denote the following statement for $ngeq 1$:
$$
S(n) : 1+r+r^2+cdots+r^n=frac{r^{n+1}-1}{r-1}.
$$
In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $ngeq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $rneq 0$ without loss of generality.



Base step ($n=1$): The statement $S(1)$ says that $1+r=frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $rneq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=frac{r-1}{r-1}$.]



Inductive step ($S(k)to S(k+1)$): Fix some $kgeq 1$ and suppose that $S(k)$ is true; that is, assume that
$$
S(k) : 1+r+r^2+cdots+r^k=frac{r^{k+1}-1}{r-1}
$$
is true. We must show that $S(k+1)$ follows where
$$
S(k+1) : 1+r+r^2+cdots+r^k+r^{k+1}=frac{r^{k+2}-1}{r-1}.
$$
Starting with the left-hand side of $S(k+1)$,
begin{align}
text{LHS} &= color{red}{1+r+r^2+cdots+r^k}+r^{k+1}tag{by definition}\[1em]
&= color{red}{frac{r^{k+1}-1}{r-1}}+r^{k+1}tag{by $S(k)$}\[1em]
&= frac{r^{k+1}-1}{r-1}+frac{r^{k+1}(r-1)}{r-1}tag{common denom.}\[1em]
&= frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}tag{combine like terms}\[1em]
&= frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}tag{expand}\[1em]
&= frac{r^{k+2}-1}{r-1},tag{simplify}
end{align}
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)to S(k+1)$, completing the inductive step.



By mathematical induction, for all $ngeq 1, S(n)$ holds true. $blacksquare$




Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.






share|cite|improve this answer











$endgroup$





















    -1












    $begingroup$

    $$1+r=frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
    I don't need to tell you more.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why this downvote ?
      $endgroup$
      – Yves Daoust
      Mar 23 '15 at 9:11











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f896832%2fproving-the-geometric-series-sum-i-0n-ri-frac1-rn11-r-by-induc%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Note: First note that you are really summing the geometric series ($a,rinmathbb{R}, aneq 0, rneq 1$)
    $$
    a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}=afrac{1-r^{n+1}}{1-r}
    $$
    with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.



    Claim: Let $a$ and $r$ be real numbers with $aneq 0$ and $rneq 1$. Prove that for each integer $ngeq 1$ that
    $$
    a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}.
    $$
    Proof. Fix $rinmathbb{R}, rneq 1$, and let $S(n)$ denote the following statement for $ngeq 1$:
    $$
    S(n) : 1+r+r^2+cdots+r^n=frac{r^{n+1}-1}{r-1}.
    $$
    In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $ngeq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $rneq 0$ without loss of generality.



    Base step ($n=1$): The statement $S(1)$ says that $1+r=frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $rneq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=frac{r-1}{r-1}$.]



    Inductive step ($S(k)to S(k+1)$): Fix some $kgeq 1$ and suppose that $S(k)$ is true; that is, assume that
    $$
    S(k) : 1+r+r^2+cdots+r^k=frac{r^{k+1}-1}{r-1}
    $$
    is true. We must show that $S(k+1)$ follows where
    $$
    S(k+1) : 1+r+r^2+cdots+r^k+r^{k+1}=frac{r^{k+2}-1}{r-1}.
    $$
    Starting with the left-hand side of $S(k+1)$,
    begin{align}
    text{LHS} &= color{red}{1+r+r^2+cdots+r^k}+r^{k+1}tag{by definition}\[1em]
    &= color{red}{frac{r^{k+1}-1}{r-1}}+r^{k+1}tag{by $S(k)$}\[1em]
    &= frac{r^{k+1}-1}{r-1}+frac{r^{k+1}(r-1)}{r-1}tag{common denom.}\[1em]
    &= frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}tag{combine like terms}\[1em]
    &= frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}tag{expand}\[1em]
    &= frac{r^{k+2}-1}{r-1},tag{simplify}
    end{align}
    we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)to S(k+1)$, completing the inductive step.



    By mathematical induction, for all $ngeq 1, S(n)$ holds true. $blacksquare$




    Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Note: First note that you are really summing the geometric series ($a,rinmathbb{R}, aneq 0, rneq 1$)
      $$
      a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}=afrac{1-r^{n+1}}{1-r}
      $$
      with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.



      Claim: Let $a$ and $r$ be real numbers with $aneq 0$ and $rneq 1$. Prove that for each integer $ngeq 1$ that
      $$
      a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}.
      $$
      Proof. Fix $rinmathbb{R}, rneq 1$, and let $S(n)$ denote the following statement for $ngeq 1$:
      $$
      S(n) : 1+r+r^2+cdots+r^n=frac{r^{n+1}-1}{r-1}.
      $$
      In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $ngeq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $rneq 0$ without loss of generality.



      Base step ($n=1$): The statement $S(1)$ says that $1+r=frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $rneq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=frac{r-1}{r-1}$.]



      Inductive step ($S(k)to S(k+1)$): Fix some $kgeq 1$ and suppose that $S(k)$ is true; that is, assume that
      $$
      S(k) : 1+r+r^2+cdots+r^k=frac{r^{k+1}-1}{r-1}
      $$
      is true. We must show that $S(k+1)$ follows where
      $$
      S(k+1) : 1+r+r^2+cdots+r^k+r^{k+1}=frac{r^{k+2}-1}{r-1}.
      $$
      Starting with the left-hand side of $S(k+1)$,
      begin{align}
      text{LHS} &= color{red}{1+r+r^2+cdots+r^k}+r^{k+1}tag{by definition}\[1em]
      &= color{red}{frac{r^{k+1}-1}{r-1}}+r^{k+1}tag{by $S(k)$}\[1em]
      &= frac{r^{k+1}-1}{r-1}+frac{r^{k+1}(r-1)}{r-1}tag{common denom.}\[1em]
      &= frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}tag{combine like terms}\[1em]
      &= frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}tag{expand}\[1em]
      &= frac{r^{k+2}-1}{r-1},tag{simplify}
      end{align}
      we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)to S(k+1)$, completing the inductive step.



      By mathematical induction, for all $ngeq 1, S(n)$ holds true. $blacksquare$




      Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Note: First note that you are really summing the geometric series ($a,rinmathbb{R}, aneq 0, rneq 1$)
        $$
        a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}=afrac{1-r^{n+1}}{1-r}
        $$
        with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.



        Claim: Let $a$ and $r$ be real numbers with $aneq 0$ and $rneq 1$. Prove that for each integer $ngeq 1$ that
        $$
        a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}.
        $$
        Proof. Fix $rinmathbb{R}, rneq 1$, and let $S(n)$ denote the following statement for $ngeq 1$:
        $$
        S(n) : 1+r+r^2+cdots+r^n=frac{r^{n+1}-1}{r-1}.
        $$
        In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $ngeq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $rneq 0$ without loss of generality.



        Base step ($n=1$): The statement $S(1)$ says that $1+r=frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $rneq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=frac{r-1}{r-1}$.]



        Inductive step ($S(k)to S(k+1)$): Fix some $kgeq 1$ and suppose that $S(k)$ is true; that is, assume that
        $$
        S(k) : 1+r+r^2+cdots+r^k=frac{r^{k+1}-1}{r-1}
        $$
        is true. We must show that $S(k+1)$ follows where
        $$
        S(k+1) : 1+r+r^2+cdots+r^k+r^{k+1}=frac{r^{k+2}-1}{r-1}.
        $$
        Starting with the left-hand side of $S(k+1)$,
        begin{align}
        text{LHS} &= color{red}{1+r+r^2+cdots+r^k}+r^{k+1}tag{by definition}\[1em]
        &= color{red}{frac{r^{k+1}-1}{r-1}}+r^{k+1}tag{by $S(k)$}\[1em]
        &= frac{r^{k+1}-1}{r-1}+frac{r^{k+1}(r-1)}{r-1}tag{common denom.}\[1em]
        &= frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}tag{combine like terms}\[1em]
        &= frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}tag{expand}\[1em]
        &= frac{r^{k+2}-1}{r-1},tag{simplify}
        end{align}
        we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)to S(k+1)$, completing the inductive step.



        By mathematical induction, for all $ngeq 1, S(n)$ holds true. $blacksquare$




        Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.






        share|cite|improve this answer











        $endgroup$



        Note: First note that you are really summing the geometric series ($a,rinmathbb{R}, aneq 0, rneq 1$)
        $$
        a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}=afrac{1-r^{n+1}}{1-r}
        $$
        with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.



        Claim: Let $a$ and $r$ be real numbers with $aneq 0$ and $rneq 1$. Prove that for each integer $ngeq 1$ that
        $$
        a+ar+ar^2+cdots+ar^n=afrac{r^{n+1}-1}{r-1}.
        $$
        Proof. Fix $rinmathbb{R}, rneq 1$, and let $S(n)$ denote the following statement for $ngeq 1$:
        $$
        S(n) : 1+r+r^2+cdots+r^n=frac{r^{n+1}-1}{r-1}.
        $$
        In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $ngeq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $rneq 0$ without loss of generality.



        Base step ($n=1$): The statement $S(1)$ says that $1+r=frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $rneq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=frac{r-1}{r-1}$.]



        Inductive step ($S(k)to S(k+1)$): Fix some $kgeq 1$ and suppose that $S(k)$ is true; that is, assume that
        $$
        S(k) : 1+r+r^2+cdots+r^k=frac{r^{k+1}-1}{r-1}
        $$
        is true. We must show that $S(k+1)$ follows where
        $$
        S(k+1) : 1+r+r^2+cdots+r^k+r^{k+1}=frac{r^{k+2}-1}{r-1}.
        $$
        Starting with the left-hand side of $S(k+1)$,
        begin{align}
        text{LHS} &= color{red}{1+r+r^2+cdots+r^k}+r^{k+1}tag{by definition}\[1em]
        &= color{red}{frac{r^{k+1}-1}{r-1}}+r^{k+1}tag{by $S(k)$}\[1em]
        &= frac{r^{k+1}-1}{r-1}+frac{r^{k+1}(r-1)}{r-1}tag{common denom.}\[1em]
        &= frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}tag{combine like terms}\[1em]
        &= frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}tag{expand}\[1em]
        &= frac{r^{k+2}-1}{r-1},tag{simplify}
        end{align}
        we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)to S(k+1)$, completing the inductive step.



        By mathematical induction, for all $ngeq 1, S(n)$ holds true. $blacksquare$




        Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:19









        Community♦

        1




        1










        answered Mar 23 '15 at 9:01









        Daniel W. FarlowDaniel W. Farlow

        17.6k114488




        17.6k114488























            -1












            $begingroup$

            $$1+r=frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
            I don't need to tell you more.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why this downvote ?
              $endgroup$
              – Yves Daoust
              Mar 23 '15 at 9:11
















            -1












            $begingroup$

            $$1+r=frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
            I don't need to tell you more.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why this downvote ?
              $endgroup$
              – Yves Daoust
              Mar 23 '15 at 9:11














            -1












            -1








            -1





            $begingroup$

            $$1+r=frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
            I don't need to tell you more.






            share|cite|improve this answer









            $endgroup$



            $$1+r=frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
            I don't need to tell you more.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 23 '15 at 9:10









            Yves DaoustYves Daoust

            127k673226




            127k673226












            • $begingroup$
              Why this downvote ?
              $endgroup$
              – Yves Daoust
              Mar 23 '15 at 9:11


















            • $begingroup$
              Why this downvote ?
              $endgroup$
              – Yves Daoust
              Mar 23 '15 at 9:11
















            $begingroup$
            Why this downvote ?
            $endgroup$
            – Yves Daoust
            Mar 23 '15 at 9:11




            $begingroup$
            Why this downvote ?
            $endgroup$
            – Yves Daoust
            Mar 23 '15 at 9:11


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f896832%2fproving-the-geometric-series-sum-i-0n-ri-frac1-rn11-r-by-induc%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more