The expression ( 1 + q ) ( 1 + q 2 ) ( 1 + q 4 ) ( 1 + q 8 ) ( 1 + q 16 ) ( 1 + q 32 ) ( 1 + q 64 ) where q ≠ 1, equals(A) 1 − q 128 1 − q (B) 1 − q 64 1 − q (C) 1 − q 2 1 + 2 + ⋯ + 6 1 − q (D) none of the foregoing expressionsWhat I have done ( 1 + q ) ( 1 + q 2 ) ( 1 + q 4 ) ( 1 + q 8 ) ( 1 + q 16 ) ( 1 + q 32 ) ( 1 + q 64 ) is a polynomial of degree 127. Now the highest degree of the polynomial in option (A), (B) and (C) is 127, 63 and 41 respectively. And therefore (A) is the correct answer.I get to the correct answer but I don't think that my way of doing is correct. I mean what if, if option (B) was 1 + q 128 1 − q Please show how should I approach to the problem to get to the correct answer without any confusion.

Question

The expression   (  1  +  q  )  (  1  +      q    2    )  (  1  +      q    4    )  (  1  +      q    8    )  (  1  +      q          16        )  (  1  +      q          32        )  (  1  +      q          64        ) where   q  ≠  1, equals(A)             1      −              q                  128                            1      −      q      (B)             1      −              q                  64                            1      −      q      (C)             1      −              q                              2                          1              +              2              +              ⋯              +              6                                                  1      −      q      (D) none of the foregoing expressionsWhat I have done  (  1  +  q  )  (  1  +      q    2    )  (  1  +      q    4    )  (  1  +      q    8    )  (  1  +      q          16        )  (  1  +      q          32        )  (  1  +      q          64        ) is a polynomial of degree 127. Now the highest degree of the polynomial in option (A), (B) and (C) is 127, 63 and 41 respectively. And therefore (A) is the correct answer.I get to the correct answer but I don&#039;t think that my way of doing is correct. I mean what if, if option (B) was             1      +              q                  128                            1      −      q      Please show how should I approach to the problem to get to the correct answer without any confusion.

zepplinkid7yk · Accepted Answer

(  1  +  q  )  (  1  +      q    2    )  =  1  +  q  +      q    2    +      q    3    =            1      −              q        4                    1      −      q      So,  (  1  +  q  )  (  1  +      q    2    )  (  1  +      q    4    )  =            (      1      −              q        4            )      (      1      +              q        4            )              1      −      q        =            1      −              q        8                    1      −      q      So,  (  1  +  q  )  (  1  +      q    2    )  (  1  +      q    4    )  (  1  +      q    8    )  =            (      1      −              q        8            )      (      1      +              q        8            )              1      −      q        =            1      −              q                  16                            1      −      q      Can you see the pattern?

cyfwelestoi · Accepted Answer

Let  P  =  (  1  +  q  )  (  1  +      q    2    )  (  1  +      q    4    )  (  1  +      q    8    )  (  1  +      q          16        )  (  1  +      q          32        )  (  1  +      q          64        )  .One has  (  1  −  q  )  P  =  (  1  −  q  )  (  1  +  q  )  (  1  +      q    2    )  (  1  +      q    4    )  (  1  +      q    8    )  (  1  +      q          16        )  (  1  +      q          32        )  (  1  +      q          64        )  =  1  −      q          128        .So, one gets  P  =            1      −              q                  128                            1      −      q

The expression ( 1 + q ) ( 1 + q 2 </msup> ) ( 1

Answered question

Answer & Explanation

New Questions in Pre-Algebra