A proof of a sum notation using induction - Need a hint.

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Heres a problem from the induction / sum notation part of the book im studying from. Can anyone hint me about how to prove it?



Show that for every $x,y$ :
$$ (x^n-y^n) = (x-y)sum_i=1^n x^n-i cdot y^i-1$$



I tried to change the border of the sum to be i=n to 2n, and then use induction but it didn't seem to work.



Thanks.




algebra-precalculus induction




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  • How about rewriting it as $$x^n=y^n+(x-y)sum_i=1^nx^n-iy^i-1?$$
    – Lord Shark the Unknown
    Jul 27 at 11:04














up vote
0
down vote

favorite












Heres a problem from the induction / sum notation part of the book im studying from. Can anyone hint me about how to prove it?



Show that for every $x,y$ :
$$ (x^n-y^n) = (x-y)sum_i=1^n x^n-i cdot y^i-1$$



I tried to change the border of the sum to be i=n to 2n, and then use induction but it didn't seem to work.



Thanks.




algebra-precalculus induction




share|cite|improve this question



















  • How about rewriting it as $$x^n=y^n+(x-y)sum_i=1^nx^n-iy^i-1?$$
    – Lord Shark the Unknown
    Jul 27 at 11:04












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Heres a problem from the induction / sum notation part of the book im studying from. Can anyone hint me about how to prove it?



Show that for every $x,y$ :
$$ (x^n-y^n) = (x-y)sum_i=1^n x^n-i cdot y^i-1$$



I tried to change the border of the sum to be i=n to 2n, and then use induction but it didn't seem to work.



Thanks.




algebra-precalculus induction




share|cite|improve this question











Heres a problem from the induction / sum notation part of the book im studying from. Can anyone hint me about how to prove it?



Show that for every $x,y$ :
$$ (x^n-y^n) = (x-y)sum_i=1^n x^n-i cdot y^i-1$$



I tried to change the border of the sum to be i=n to 2n, and then use induction but it didn't seem to work.



Thanks.





algebra-precalculus induction





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share|cite|improve this question




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asked Jul 27 at 11:02









Dvir Peretz

526




526











  • How about rewriting it as $$x^n=y^n+(x-y)sum_i=1^nx^n-iy^i-1?$$
    – Lord Shark the Unknown
    Jul 27 at 11:04
















  • How about rewriting it as $$x^n=y^n+(x-y)sum_i=1^nx^n-iy^i-1?$$
    – Lord Shark the Unknown
    Jul 27 at 11:04















How about rewriting it as $$x^n=y^n+(x-y)sum_i=1^nx^n-iy^i-1?$$
– Lord Shark the Unknown
Jul 27 at 11:04




How about rewriting it as $$x^n=y^n+(x-y)sum_i=1^nx^n-iy^i-1?$$
– Lord Shark the Unknown
Jul 27 at 11:04










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










If $n=1$, then all that the equality asserts is that $x-y=x-y$. Now, assuming that the equaliy holds for a certain natural $n$ (and for every $x$ and evey $y$), you can observe thatbeginalignx^n+1-y^n+1&=x^n+1-x^ny+x^ny-y^n+1\&=x^n(x-y)+(x^n-y^n)yendalignand then you can use the induction hypothesis.



On the other hand, I would write the sum $displaystylesum_i=1^nx^n-iy^i-1$ as $x^n-1+x^n-2y+cdots+xy^n-2+y^n-1$.






share|cite|improve this answer





















  • I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
    – Dvir Peretz
    Jul 27 at 11:23







  • 1




    I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
    – José Carlos Santos
    Jul 27 at 11:36

















up vote
0
down vote













A proof using induction always has two parts. Naturally, it starts with a statement about all integers, i.e. it has the form $forall ninmathbb N: P(n)$



  1. In the first part, you must prove that the statement is true for the integer $1$. In other words, you must prove $P(1)$.

  2. In the second part, you can assume that the statement is true for $n$ (where $n$ is general), and from that, you must prove that the statement is true for $n+1$. In other words, you must prove $P(n)implies P(n+1)$.

I don't know exactly where you are stuck, so please, in the comment, answer these questions:



  • Did you write down what the statement looks like for $n=1$? If so, what does it look like?

  • Did you already prove the statement for $n=1$?

  • Did you write down what the statement looks like for $n+1$? If so, what does it look like?

After you answer these, I can help you further.






share|cite|improve this answer





















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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    If $n=1$, then all that the equality asserts is that $x-y=x-y$. Now, assuming that the equaliy holds for a certain natural $n$ (and for every $x$ and evey $y$), you can observe thatbeginalignx^n+1-y^n+1&=x^n+1-x^ny+x^ny-y^n+1\&=x^n(x-y)+(x^n-y^n)yendalignand then you can use the induction hypothesis.



    On the other hand, I would write the sum $displaystylesum_i=1^nx^n-iy^i-1$ as $x^n-1+x^n-2y+cdots+xy^n-2+y^n-1$.






    share|cite|improve this answer





















    • I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
      – Dvir Peretz
      Jul 27 at 11:23







    • 1




      I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
      – José Carlos Santos
      Jul 27 at 11:36














    up vote
    0
    down vote



    accepted










    If $n=1$, then all that the equality asserts is that $x-y=x-y$. Now, assuming that the equaliy holds for a certain natural $n$ (and for every $x$ and evey $y$), you can observe thatbeginalignx^n+1-y^n+1&=x^n+1-x^ny+x^ny-y^n+1\&=x^n(x-y)+(x^n-y^n)yendalignand then you can use the induction hypothesis.



    On the other hand, I would write the sum $displaystylesum_i=1^nx^n-iy^i-1$ as $x^n-1+x^n-2y+cdots+xy^n-2+y^n-1$.






    share|cite|improve this answer





















    • I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
      – Dvir Peretz
      Jul 27 at 11:23







    • 1




      I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
      – José Carlos Santos
      Jul 27 at 11:36












    up vote
    0
    down vote



    accepted







    up vote
    0
    down vote



    accepted






    If $n=1$, then all that the equality asserts is that $x-y=x-y$. Now, assuming that the equaliy holds for a certain natural $n$ (and for every $x$ and evey $y$), you can observe thatbeginalignx^n+1-y^n+1&=x^n+1-x^ny+x^ny-y^n+1\&=x^n(x-y)+(x^n-y^n)yendalignand then you can use the induction hypothesis.



    On the other hand, I would write the sum $displaystylesum_i=1^nx^n-iy^i-1$ as $x^n-1+x^n-2y+cdots+xy^n-2+y^n-1$.






    share|cite|improve this answer













    If $n=1$, then all that the equality asserts is that $x-y=x-y$. Now, assuming that the equaliy holds for a certain natural $n$ (and for every $x$ and evey $y$), you can observe thatbeginalignx^n+1-y^n+1&=x^n+1-x^ny+x^ny-y^n+1\&=x^n(x-y)+(x^n-y^n)yendalignand then you can use the induction hypothesis.



    On the other hand, I would write the sum $displaystylesum_i=1^nx^n-iy^i-1$ as $x^n-1+x^n-2y+cdots+xy^n-2+y^n-1$.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 27 at 11:08









    José Carlos Santos

    113k1696173




    113k1696173











    • I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
      – Dvir Peretz
      Jul 27 at 11:23







    • 1




      I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
      – José Carlos Santos
      Jul 27 at 11:36
















    • I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
      – Dvir Peretz
      Jul 27 at 11:23







    • 1




      I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
      – José Carlos Santos
      Jul 27 at 11:36















    I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
    – Dvir Peretz
    Jul 27 at 11:23





    I used the induction notation, (Switching $x^n-y^n$ for the sum part, Then took out (x-y) and changed the borders & put $x^ny$ and $y$ into the sum.) Thank you very much, But i don't think i would figure that by my self. Is that a technique for solving sum notation induction problems?
    – Dvir Peretz
    Jul 27 at 11:23





    1




    1




    I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
    – José Carlos Santos
    Jul 27 at 11:36




    I had to find a way to use the induction hypothesis. In particular, I had to find a way to express $x^n+1-y^n+1$ that used the expression $x^n-y^n$.
    – José Carlos Santos
    Jul 27 at 11:36










    up vote
    0
    down vote













    A proof using induction always has two parts. Naturally, it starts with a statement about all integers, i.e. it has the form $forall ninmathbb N: P(n)$



    1. In the first part, you must prove that the statement is true for the integer $1$. In other words, you must prove $P(1)$.

    2. In the second part, you can assume that the statement is true for $n$ (where $n$ is general), and from that, you must prove that the statement is true for $n+1$. In other words, you must prove $P(n)implies P(n+1)$.

    I don't know exactly where you are stuck, so please, in the comment, answer these questions:



    • Did you write down what the statement looks like for $n=1$? If so, what does it look like?

    • Did you already prove the statement for $n=1$?

    • Did you write down what the statement looks like for $n+1$? If so, what does it look like?

    After you answer these, I can help you further.






    share|cite|improve this answer

























      up vote
      0
      down vote













      A proof using induction always has two parts. Naturally, it starts with a statement about all integers, i.e. it has the form $forall ninmathbb N: P(n)$



      1. In the first part, you must prove that the statement is true for the integer $1$. In other words, you must prove $P(1)$.

      2. In the second part, you can assume that the statement is true for $n$ (where $n$ is general), and from that, you must prove that the statement is true for $n+1$. In other words, you must prove $P(n)implies P(n+1)$.

      I don't know exactly where you are stuck, so please, in the comment, answer these questions:



      • Did you write down what the statement looks like for $n=1$? If so, what does it look like?

      • Did you already prove the statement for $n=1$?

      • Did you write down what the statement looks like for $n+1$? If so, what does it look like?

      After you answer these, I can help you further.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        A proof using induction always has two parts. Naturally, it starts with a statement about all integers, i.e. it has the form $forall ninmathbb N: P(n)$



        1. In the first part, you must prove that the statement is true for the integer $1$. In other words, you must prove $P(1)$.

        2. In the second part, you can assume that the statement is true for $n$ (where $n$ is general), and from that, you must prove that the statement is true for $n+1$. In other words, you must prove $P(n)implies P(n+1)$.

        I don't know exactly where you are stuck, so please, in the comment, answer these questions:



        • Did you write down what the statement looks like for $n=1$? If so, what does it look like?

        • Did you already prove the statement for $n=1$?

        • Did you write down what the statement looks like for $n+1$? If so, what does it look like?

        After you answer these, I can help you further.






        share|cite|improve this answer













        A proof using induction always has two parts. Naturally, it starts with a statement about all integers, i.e. it has the form $forall ninmathbb N: P(n)$



        1. In the first part, you must prove that the statement is true for the integer $1$. In other words, you must prove $P(1)$.

        2. In the second part, you can assume that the statement is true for $n$ (where $n$ is general), and from that, you must prove that the statement is true for $n+1$. In other words, you must prove $P(n)implies P(n+1)$.

        I don't know exactly where you are stuck, so please, in the comment, answer these questions:



        • Did you write down what the statement looks like for $n=1$? If so, what does it look like?

        • Did you already prove the statement for $n=1$?

        • Did you write down what the statement looks like for $n+1$? If so, what does it look like?

        After you answer these, I can help you further.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 27 at 11:07









        5xum

        81.8k382146




        81.8k382146






















             

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