therightwomanwf

2022-07-03

A method I got from somewhere.
$\begin{array}{rl}f\left(x\right)& =x+{x}^{3}+{x}^{5}+\cdots +{x}^{2n-1}\\ & =x\left[\left({x}^{2\left(n-1\right)}-1\right)+\left({x}^{2\left(n-2\right)}-1\right)+\cdots +\left({x}^{2×1}-1\right)+\left({x}^{2×0}-1\right)+n\right]\\ & =x\left({x}^{2\left(n-1\right)}-1\right)+x\left({x}^{2\left(n-2\right)}-1\right)+\cdots +x\left({x}^{2×1}-1\right)+x\left({x}^{2×0}-1\right)+nx\end{array}$
As the last term nx cannot be divided by ${x}^{3}-x$, the remainder is nx.

Kaya Kemp

Expert

First, we can eliminate a factor of x , and find the remainder of
$P\left(x\right)=1+{x}^{2}+{x}^{4}+\cdots +{x}^{2n-2}$ divided by ${x}^{2}-1$
Write $P\left(x\right)$ as
$P\left(x\right)=\left({x}^{2}-1\right)Q\left(x\right)+ax+b$
Now substitute the values $x=1$, $x=-1$ to obtain
$P\left(1\right)=a+b$
$P\left(-1\right)=b-a$
and note that $P\left(1\right)=P\left(-1\right)=n$, and solve the simultaneous equations

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