Simplification of ∑ k = 1 n 1 4 k 2 − 1 and Wolfram Alpha tells me it can be simplified to two forms: n 2 n + 1 and 1 2 − 1 2 ( 2 n + 1 ) The problem being that I don't know how to do so and I cannot seem to, after sincerely trying, find any helpful information in my text book nor in Google. Thanks.

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Simplification of       ∑          k      =      1        n        1          4              k        2            −      1      and Wolfram Alpha tells me it can be simplified to two forms:      n          2      n      +      1      and      1    2    −      1          2      (      2      n      +      1      )      The problem being that I don&#039;t know how to do so and I cannot seem to, after sincerely trying, find any helpful information in my text book nor in Google. Thanks.

treccinair · Accepted Answer

Hint:      1          4              k        2            −      1        =      1          (      2      k      +      1      )      (      2      k      −      1      )        =      1    2        (          1              2        k        −        1              −          1              2        k        +        1              )  So now your sum can collapse to contain only some initial and terminal terms.

slijmigrd · Accepted Answer

There is a problem in what you say.Hint      1          4              k        2            −      1        =      1          (      2      k      −      1      )      (      2      k      +      1      )        .Then, find   A  ,  B such that      1          4              k        2            −      1        =      A          2      k      −      1        +      B          2      k      +      1        ,and your sum will follow...

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