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

Monfredo0n

Monfredo0n

Answered question

2022-05-19

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 expressions
What 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.

Answer & Explanation

zepplinkid7yk

zepplinkid7yk

Beginner2022-05-20Added 11 answers

( 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

cyfwelestoi

Beginner2022-05-21Added 3 answers

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

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