A method I got from somewhere. <mtable columnalign="right left right left right left right left

therightwomanwf

therightwomanwf

Answered question

2022-07-03

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.

Answer & Explanation

Kaya Kemp

Kaya Kemp

Beginner2022-07-04Added 18 answers

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 1
Write P ( x ) as
P ( x ) = ( x 2 1 ) Q ( x ) + a x + b
Now substitute the values x = 1, x = 1 to obtain
P ( 1 ) = a + b
P ( 1 ) = b a
and note that P ( 1 ) = P ( 1 ) = n, and solve the simultaneous equations

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