therightwomanwf

Answered

2022-07-03

A method I got from somewhere.

$\begin{array}{rl}f(x)& =x+{x}^{3}+{x}^{5}+\cdots +{x}^{2n-1}\\ & =x{\textstyle [}{\textstyle (}{x}^{2(n-1)}-1{\textstyle )}+{\textstyle (}{x}^{2(n-2)}-1{\textstyle )}+\cdots +{\textstyle (}{x}^{2\times 1}-1{\textstyle )}+{\textstyle (}{x}^{2\times 0}-1{\textstyle )}+n{\textstyle ]}\\ & =x{\textstyle (}{x}^{2(n-1)}-1{\textstyle )}+x{\textstyle (}{x}^{2(n-2)}-1{\textstyle )}+\cdots +x{\textstyle (}{x}^{2\times 1}-1{\textstyle )}+x{\textstyle (}{x}^{2\times 0}-1{\textstyle )}+nx\end{array}$

As the last term nx cannot be divided by ${x}^{3}-x$, the remainder is nx.

$\begin{array}{rl}f(x)& =x+{x}^{3}+{x}^{5}+\cdots +{x}^{2n-1}\\ & =x{\textstyle [}{\textstyle (}{x}^{2(n-1)}-1{\textstyle )}+{\textstyle (}{x}^{2(n-2)}-1{\textstyle )}+\cdots +{\textstyle (}{x}^{2\times 1}-1{\textstyle )}+{\textstyle (}{x}^{2\times 0}-1{\textstyle )}+n{\textstyle ]}\\ & =x{\textstyle (}{x}^{2(n-1)}-1{\textstyle )}+x{\textstyle (}{x}^{2(n-2)}-1{\textstyle )}+\cdots +x{\textstyle (}{x}^{2\times 1}-1{\textstyle )}+x{\textstyle (}{x}^{2\times 0}-1{\textstyle )}+nx\end{array}$

As the last term nx cannot be divided by ${x}^{3}-x$, the remainder is nx.

Answer & Explanation

Kaya Kemp

Expert

2022-07-04Added 18 answers

First, we can eliminate a factor of x , and find the remainder of

$P(x)=1+{x}^{2}+{x}^{4}+\cdots +{x}^{2n-2}$ divided by ${x}^{2}-1$

Write $P(x)$ as

$P(x)=({x}^{2}-1)Q(x)+ax+b$

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

$P(x)=1+{x}^{2}+{x}^{4}+\cdots +{x}^{2n-2}$ divided by ${x}^{2}-1$

Write $P(x)$ as

$P(x)=({x}^{2}-1)Q(x)+ax+b$

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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