A method I got from somewhere. f ( x ) = x + x 3 + x 5 + ⋯ + x 2 n − 1 = x [ ( x 2 ( n − 1 ) − 1 ) + ( x 2 ( n − 2 ) − 1 ) + ⋯ + ( x 2 × 1 − 1 ) + ( x 2 × 0 − 1 ) + n ] = x ( x 2 ( n − 1 ) − 1 ) + x ( x 2 ( n − 2 ) − 1 ) + ⋯ + x ( x 2 × 1 − 1 ) + x ( x 2 × 0 − 1 ) + n x As the last term nx cannot be divided by x 3 − x, the remainder is nx.

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A method I got from somewhere.                    f        (        x        )                            =        x        +                  x          3                +                  x          5                +        ⋯        +                  x                      2            n            −            1                                                            =        x                              [                                                (                                    x                      2            (            n            −            1            )                          −        1                              )                          +                              (                                    x                      2            (            n            −            2            )                          −        1                              )                          +        ⋯        +                              (                                    x                      2            ×            1                          −        1                              )                          +                              (                                    x                      2            ×            0                          −        1                              )                          +        n                              ]                                                            =        x                              (                                    x                      2            (            n            −            1            )                          −        1                              )                          +        x                              (                                    x                      2            (            n            −            2            )                          −        1                              )                          +        ⋯        +        x                              (                                    x                      2            ×            1                          −        1                              )                          +        x                              (                                    x                      2            ×            0                          −        1                              )                          +        n        x            As the last term nx cannot be divided by       x    3    −  x, the remainder is nx.

Kaya Kemp · Accepted Answer

First, we can eliminate a factor of x , and find the remainder of  P  (  x  )  =  1  +      x    2    +      x    4    +  ⋯  +      x          2      n      −      2       divided by       x    2    −  1Write   P  (  x  ) as  P  (  x  )  =  (      x    2    −  1  )  Q  (  x  )  +  a  x  +  bNow substitute the values   x  =  1,   x  =  −  1 to obtain  P  (  1  )  =  a  +  b  P  (  −  1  )  =  b  −  aand note that   P  (  1  )  =  P  (  −  1  )  =  n, and solve the simultaneous equations

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