Solve ∑ i = 1 200 1 1 + x i = ? ( x 2 + x + 1 ) 100 = a 0 + a 1 x + a 2 x 2 + . . . + a 199 x 199 + a 200 x 200 ∑ i = 1 200 1 1 + x i = ?Can somebody help me? Thank you!

Question

Solve       ∑          i      =      1              200                  1              1        +                  x          i                      =  ?  (      x          2        +  x  +  1      )          100        =      a    0    +      a    1    x  +      a    2        x          2        +  .  .  .  +      a          199            x          199        +      a          200            x          200            ∑          i      =      1              200                  1              1        +                  x          i                      =  ?Can somebody help me? Thank you!

Haylie Campbell · Accepted Answer

∑          k      =      1              200            1          1      +              x        k              =  100      (          1              1        +                  x          1                      +          1              1        +                  x          2                      )    =            100      (      2      +              x        1            +              x        2            )              1      +              x        1            +              x        2            +              x        1                    x        2              =  =            100      (      2      −      1      )              1      −      1      +      1        =  100.

(x^(2)+x+1)^(100)=a_0+a_1x+a_2x^(2)+...+a_(199)x^(199)+a_(200)x^(200) sum_(i=1)^(200) (1/(1+x_i))=? Can somebody help me? Thank you!

Answered question

Answer & Explanation

New Questions in Pre-Algebra