As a follow up of my previous question here is an improved version of the series: ∑ i = 1 2 N R i ⋅ ( 1 1 − p ) i ( 1 − 2 R − ( 2 N R − i ) Δ 2 R )

Question

As a follow up of my previous question here is an improved version of the series:      ∑          i      =      1              2      N      R            i    ⋅                  (                              1                          1              −              p                                      )            i              (    1    −                            2          R          −          (          2          N          R          −          i          )          Δ                          2          R                      )

Louis Lawrence · Accepted Answer

Note that a little algebra will transform your sum into      ∑          i      =      1              2      N      R            (    K    i    +          Δ              2        R                    i      2        )              (              1                  1          −          p                    )        i    ,where K is an easily computed constant.The answer to the previous post tells you how to find       ∑    1          2      N      R        i      x    i  . Multiply that sum by K. The only remaining task is to find       ∑    1          2      N      R            i    2        x    i  This is done using more or less the same idea as the preceding answer by Marvis. Let   f  (  x  )  =  1  +  x  +      x    2    +  ⋯  +      x    n  . There is an explicit closed form for   f  (  x  ). Note that  x      f    ′    (  x  )  =      ∑          i      =      0        n    i      x          i        .If   g  (  x  )  =  x      f    ′    (  x  ), then by the same trick  x      g    ′    (  x  )  =      ∑          i      =      0        n        i    2        x          i        .

As a follow up of my previous question here is an improved version of the series: <munderover>

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New Questions in Calculus 2