Prove that ∑ 1 ≤ k &lt; j ≤ n tan 2 ⁡ ( k π 2 n + 1 ) tan 2 ⁡ ( j π 2 n + 1 ) = ( 2 n + 1 4 )

Question

Prove that             ∑              1        ≤        k        &amp;lt;        j        ≤        n                    tan      2        ⁡          (                        k          π                          2          n          +          1                    )              tan      2        ⁡          (                        j          π                          2          n          +          1                    )        =                            (                                      2          n          +          1                4                              )

escampetaq5 · Accepted Answer

It is enough to recall Cauchy&#039;s proof of the Basel problem, relying on the identity:                    (1)                                                        (                                                          2              n              +              1                        1                                              )                                                t          n                −                                            (                                                          2              n              +              1                        3                                              )                                                t                      n            −            1                          +        …        +        (        −        1                  )          n                                                    (                                                          2              n              +              1                                      2              n              +              1                                                          )                                      =                  ∏                      k            =            1                                n                                    (          t          −                      cot            2                    ⁡                                    k              π                                      2              n              +              1                                )                    If we consider the &quot;reciprocal polynomial&quot;                    (2)                                                        (                                                          2              n              +              1                                      2              n              +              1                                                          )                                                t          n                −                                            (                                                          2              n              +              1                                      2              n              −              1                                                          )                                                t                      n            −            1                          +        …        +        (        −        1                  )          n                                                    (                                                          2              n              +              1                        1                                              )                                      =                  ∏                      k            =            1                                n                                    (          t          −                      tan            2                    ⁡                                    k              π                                      2              n              +              1                                )                    the original sum is just the second elementary symmetric polynomial of the roots,       e    2  , that by Vieta&#039;s theorem is given by the coefficient of       t          n      −      2       in the LHS of (2), so:                    (3)                              ∑                      1            ≤            j            &amp;lt;            k            ≤            n                                    tan          2                ⁡                  (                                    j              π                                      2              n              +              1                                )                          tan          2                ⁡                  (                                    k              π                                      2              n              +              1                                )                =                                            (                                                          2              n              +              1                                      2              n              −              3                                                          )                                      =                                                            (                                                                    2                n                +                1                            4                                                      )                                                        as wanted.

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