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







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












$begingroup$


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















                                -

                                Popular posts from this blog

                                "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                                Image
                                .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness.
                                Read more

                                Alcedinidae

                                Image
                                Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
                                Read more

                                RAC Tourist Trophy

                                Image
                                Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
                                Read more