Simplify $(-1)^{n-1}frac{800}{(pi^2){(2n-1)^2}}$ [closed]











up vote
1
down vote

favorite












Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help










share|cite|improve this question















closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 at 13:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
    – Matti P.
    Nov 22 at 11:42






  • 3




    Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
    – Matti P.
    Nov 22 at 11:42






  • 1




    Where is the equation?
    – William Elliot
    Nov 22 at 11:56








  • 1




    @MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
    – Barry Cipra
    Nov 22 at 11:57








  • 1




    I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
    – Barry Cipra
    Nov 22 at 12:05















up vote
1
down vote

favorite












Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help










share|cite|improve this question















closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 at 13:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
    – Matti P.
    Nov 22 at 11:42






  • 3




    Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
    – Matti P.
    Nov 22 at 11:42






  • 1




    Where is the equation?
    – William Elliot
    Nov 22 at 11:56








  • 1




    @MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
    – Barry Cipra
    Nov 22 at 11:57








  • 1




    I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
    – Barry Cipra
    Nov 22 at 12:05













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help










share|cite|improve this question















Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help







algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 12:03









Barry Cipra

58.6k653122




58.6k653122










asked Nov 22 at 11:32









S.Bull

61




61




closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 at 13:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 at 13:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
    – Matti P.
    Nov 22 at 11:42






  • 3




    Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
    – Matti P.
    Nov 22 at 11:42






  • 1




    Where is the equation?
    – William Elliot
    Nov 22 at 11:56








  • 1




    @MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
    – Barry Cipra
    Nov 22 at 11:57








  • 1




    I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
    – Barry Cipra
    Nov 22 at 12:05














  • 1




    I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
    – Matti P.
    Nov 22 at 11:42






  • 3




    Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
    – Matti P.
    Nov 22 at 11:42






  • 1




    Where is the equation?
    – William Elliot
    Nov 22 at 11:56








  • 1




    @MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
    – Barry Cipra
    Nov 22 at 11:57








  • 1




    I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
    – Barry Cipra
    Nov 22 at 12:05








1




1




I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 at 11:42




I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 at 11:42




3




3




Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 at 11:42




Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 at 11:42




1




1




Where is the equation?
– William Elliot
Nov 22 at 11:56






Where is the equation?
– William Elliot
Nov 22 at 11:56






1




1




@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 at 11:57






@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 at 11:57






1




1




I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 at 12:05




I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 at 12:05










1 Answer
1






active

oldest

votes

















up vote
1
down vote













It seems that there are two questions here:



The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'



and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'



The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum



$$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:



$$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$



And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    It seems that there are two questions here:



    The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'



    and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'



    The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum



    $$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:



    $$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$



    And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).






    share|cite|improve this answer

























      up vote
      1
      down vote













      It seems that there are two questions here:



      The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'



      and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'



      The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum



      $$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:



      $$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$



      And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        It seems that there are two questions here:



        The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'



        and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'



        The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum



        $$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:



        $$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$



        And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).






        share|cite|improve this answer












        It seems that there are two questions here:



        The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'



        and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'



        The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum



        $$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:



        $$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$



        And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 12:37









        Vincent

        3,01611228




        3,01611228















            Popular posts from this blog

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei