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






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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











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    -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









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











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    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














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        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


















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