How to write a general formula for ${-2/9, 3/16, -4/25, 5/36…}$? [on hold]












0












$begingroup$


I'm having a bit of trouble with this sequence:




${-2/9, 3/16, -4/25, 5/36...}$




I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change






sequences-and-series algebra-precalculus







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$endgroup$



put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel

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












  • 3




    $begingroup$
    How about $(-1)^n$?
    $endgroup$
    – Pebeto
    2 days ago












  • $begingroup$
    ah makes sense. thanks!
    $endgroup$
    – vmahajan17
    2 days ago
















0












$begingroup$


I'm having a bit of trouble with this sequence:




${-2/9, 3/16, -4/25, 5/36...}$




I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change






sequences-and-series algebra-precalculus







share|cite|improve this question











$endgroup$



put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel

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












  • 3




    $begingroup$
    How about $(-1)^n$?
    $endgroup$
    – Pebeto
    2 days ago












  • $begingroup$
    ah makes sense. thanks!
    $endgroup$
    – vmahajan17
    2 days ago














0












0








0


1



$begingroup$


I'm having a bit of trouble with this sequence:




${-2/9, 3/16, -4/25, 5/36...}$




I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change






sequences-and-series algebra-precalculus







share|cite|improve this question











$endgroup$




I'm having a bit of trouble with this sequence:




${-2/9, 3/16, -4/25, 5/36...}$




I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change






sequences-and-series algebra-precalculus




sequences-and-series algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









user21820

39.4k543155




39.4k543155










asked 2 days ago









vmahajan17vmahajan17

218




218




put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel

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







put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel

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








  • 3




    $begingroup$
    How about $(-1)^n$?
    $endgroup$
    – Pebeto
    2 days ago












  • $begingroup$
    ah makes sense. thanks!
    $endgroup$
    – vmahajan17
    2 days ago














  • 3




    $begingroup$
    How about $(-1)^n$?
    $endgroup$
    – Pebeto
    2 days ago












  • $begingroup$
    ah makes sense. thanks!
    $endgroup$
    – vmahajan17
    2 days ago








3




3




$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
2 days ago






$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
2 days ago














$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
2 days ago




$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
2 days ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

Answer is $$(-1)^nfrac{n+1}{(n+2)^2}$$






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Here is the way:
    $$
    a_n=(-1)^nfrac{n+1}{(n+2)^2}
    $$






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      $$T_n=(-1)^nfrac{n+1}{(n+2^2)}$$






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        Answer is $$(-1)^nfrac{n+1}{(n+2)^2}$$






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Answer is $$(-1)^nfrac{n+1}{(n+2)^2}$$






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Answer is $$(-1)^nfrac{n+1}{(n+2)^2}$$






            share|cite|improve this answer









            $endgroup$



            Answer is $$(-1)^nfrac{n+1}{(n+2)^2}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Sujit BhattacharyyaSujit Bhattacharyya

            1,533519




            1,533519























                6












                $begingroup$

                Here is the way:
                $$
                a_n=(-1)^nfrac{n+1}{(n+2)^2}
                $$






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  Here is the way:
                  $$
                  a_n=(-1)^nfrac{n+1}{(n+2)^2}
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    Here is the way:
                    $$
                    a_n=(-1)^nfrac{n+1}{(n+2)^2}
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    Here is the way:
                    $$
                    a_n=(-1)^nfrac{n+1}{(n+2)^2}
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    Holding ArthurHolding Arthur

                    1,271417




                    1,271417























                        3












                        $begingroup$

                        $$T_n=(-1)^nfrac{n+1}{(n+2^2)}$$






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          $$T_n=(-1)^nfrac{n+1}{(n+2^2)}$$






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            $$T_n=(-1)^nfrac{n+1}{(n+2^2)}$$






                            share|cite|improve this answer









                            $endgroup$



                            $$T_n=(-1)^nfrac{n+1}{(n+2^2)}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            saket kumarsaket kumar

                            198113




                            198113















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