Find the value of x 1 2 1 − x 1 + x 2 2 1 − x 2 + . . . + x n 2 1 − x n ..Suppose, x 1 + x 2 + . . . + x n = 1 ( x i ∈ R , x i ≠ 1 ) and x 1 1 − x 1 + x 2 1 − x 2 + . . . + x n 1 − x n = 1. Find the value of x 1 2 1 − x 1 + x 2 2 1 − x 2 + . . . + x n 2 1 − x n .

Question

Find the value of             x      1      2              1      −              x        1              +            x      2      2              1      −              x        2              +  .  .  .  +            x      n      2              1      −              x        n              ..Suppose,       x    1    +      x    2    +  .  .  .  +      x    n    =  1     (      x    i    ∈      R    ,      x    i    ≠  1  ) and             x      1              1      −              x        1              +            x      2              1      −              x        2              +  .  .  .  +            x      n              1      −              x        n              =  1. Find the value of             x      1      2              1      −              x        1              +            x      2      2              1      −              x        2              +  .  .  .  +            x      n      2              1      −              x        n              .

candelo6a · Accepted Answer

x      1      2              1      −              x        1              +            x      2      2              1      −              x        2              +  .  .  .  +            x      n      2              1      −              x        n              =            x      1      2              1      −              x        1              +            x      2      2              1      −              x        2              +  .  .  .  +            x      n      2              1      −              x        n              −  1  +  1    =                    x        1        2            −              x        1                    1      −              x        1              +                    x        2        2            −              x        2                    1      −              x        2              +  .  .  .  +                    x        n        2            −              x        n                    1      −              x        n              +  1  =  −  (      x    1    +      x    2    +  .  .  +      x    n    )  +  1  =  0

telegrafyx · Accepted Answer

Let   S be the sum of the series                     x        1        2                    1        −                  x          1                      +                    x        2        2                    1        −                  x          2                      +  ⋯  +                    x        n        2                    1        −                  x          n                     , represented by   S  =      ∑          i      =      1        n                      x        i        2                    1        −                  x          i                    First, let&#039;s do some algebra.                                                        x              i              2                                      1              −                              x                i                                                                        =                                                            x                i                2                            −              2                              x                i                            +              2                              x                i                            −              1              +              1                                      1              −                              x                i                                                                                      =                                            (                              x                i                2                            −              2                              x                i                            +              1              )              +              (                              x                i                            −              1              )              +                              x                i                                                    1              −                              x                i                                                                                      =                                            (                              x                i                            −              1                              )                2                                                    1              −                              x                i                                                    +                                                            x                i                            −              1                                      1              −                              x                i                                                    +                                            x              i                                      1              −                              x                i                                                                                      =        (        1        −                  x          i                )        −        1        +                                            x              i                                      1              −                              x                i                                                                                      =                                            x              i                                      1              −                              x                i                                                    −                  x          i                    Now we can simplify our sum.  S  =      ∑          i      =      1        n                      x        i        2                    1        −                  x          i                      =      ∑          i      =      1        n    (                    x        i                    1        −                  x          i                      −      x    i    )  =      ∑          i      =      1        n                      x        i                    1        −                  x          i                      −      ∑          i      =      1        n        x    i    =  1  −  1  =  0

Find the value of <msubsup> x 1 2 </msubsup> 1

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