∑ k = 1 2004 1 1 + tan 2 ⁡ ( k π 2 ⋅ 2005 ) So I've been working on this problem for a while and I was able to get up to here: ∑ k = 1 2004 cos 2 ⁡ ( k π 2 ⋅ 2005 ) With the trigonometric identity that 1 + tan 2 ⁡ ( θ ) = sec 2 ⁡ ( θ ) . And that cos ⁡ ( θ ) = 1 sec ⁡ ( θ ) .I'm stuck at this point, does anyone have anything on how to move it forward?

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∑          k      =      1              2004            1          1      +              tan                  2                    ⁡              (                              k            π                                2            ⋅            2005                          )            So I&#039;ve been working on this problem for a while and I was able to get up to here:      ∑          k      =      1              2004            cos          2        ⁡      (                  k        π                    2        ⋅        2005              )  With the trigonometric identity that   1  +      tan    2    ⁡      (    θ    )    =      sec    2    ⁡      (    θ    )  . And that   cos  ⁡      (    θ    )    =      1          sec      ⁡              (        θ        )            .I&#039;m stuck at this point, does anyone have anything on how to move it forward?

jmibanezla · Accepted Answer

∑                      k            =            1                                2004                                    cos                      2                          ⁡                  (                                    k              π                                      2              ⋅              2005                                )                                    =                  1          2                          ∑                      k            =            1                                2004                          (        1        +        cos        ⁡                              k            π                    2005                )                                          =                  2004          2                +                  1          2                          ∑                      k            =            1                                1002                                    [          cos          ⁡                                    k              π                        2005                    +          cos          ⁡                                    (              2005              −              k              )              π                        2005                    ]                                                  =        1002        +                  1          2                          ∑                      k            =            1                                1002                          0        =        1002

<munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> k = 1

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