The evaluation, ∑ n = 0 ∞ 1 F 2 n = 7 − 5 2 = ( 1 − 5 2 ) 3 + ( 1 + 5 2 ) 2

Question

The evaluation,      ∑          n      =      0        ∞        1          F                        2          n                      =            7      −              5              2    =            (                        1          −                      5                          2            )        3    +            (                        1          +                      5                          2            )        2

Donavan Mack · Accepted Answer

Proof: Let us write out a few terms of this sequence, we get      a    0    =  0  ,      a    1    =  1  ,      a    2    =  b  ,      a    3    =      b    2    +  1  ,      a    4    =      b    3    +  2  b  ,  ⋯The proof is by induction on N. For N=1, we have the left hand side to be            1              a        1              +            1              a        2              =  1  +            1      b      while the right hand side is  1  +            2      b        −                    a        1                    a        2              =  1  +            2      b        −            1      b        =  1  +            1      b      For N=2, we have the left hand side to be            1              a        1              +            1              a        2              +            1              a        4              =  1  +            1      b        +            1                        b          3                +        2        b            while the right hand side is  1  +            2      b        −                    a        3                    a        4              =  1  +            2      b        −                              b          2                +        1                              b          3                +        2        b              =  1  +            1      b        +            1      b        −                              b          2                +        1                              b          3                +        2        b              =  1  +            1      b        +            1                        b          3                +        2        b            Hence, it holds for N=1 and N=2. Now lets go ahead with induction now. Assume the result is true for N=m i.e. we have      ∑          k      =      0              m                  1              a                              2            k                                =  1  +            2      b        −                    a                              2            m                    −          1                            a                              2            m                              Now      ∑          k      =      0              m      +      1                  1              a                              2            k                                =  1  +            2      b        −                    a                              2            m                    −          1                            a                              2            m                                +            1              a                              2                          m              +              1                                          Hence, we want to show that  −                    a                              2            m                    −          1                            a                              2            m                                +            1              a                              2                          m              +              1                                            =  −                    a                              2                          m              +              1                                −          1                            a                              2                          m              +              1                                          i.e.            1              a                              2                          m              +              1                                            +                    a                              2                          m              +              1                                −          1                            a                              2                          m              +              1                                            =                    a                              2            m                    −          1                            a                              2            m                              i.e.      a                  2        m              (  1  +      a                  2                  m          +          1                    −      1        )  =      a                  2        m            −      1            a                  2                  m          +          1                              (  ⋆  )which can be verified using the recurrence. In fact   (  †  ), a slightly more general version of   (  ⋆  ), which is easier to check is true.      a          2      k        (  1  +      a          4      k      −      1        )  =      a          2      k      −      1            a          4      k                (  †  )i.e.      a          2      k      −      1            a          4      k        −      a          2      k            a          4      k      −      1        =      a          2      k                (  †  )Hence, we get that                                                                        ∑                                  k                  =                  0                                                  N                                                                              1                                      a                                                                  2                        k                                                                                                        =              1              +                                                2                  b                                            −                                                                    a                                                                  2                        N                                            −                      1                                                                            a                                                                  2                        N

The evaluation, <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n

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