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How do I show that the following is true?

$$ A(j+x+x^2+....+x^n)= A(j-x{^{n+1}})(j-x){^{-1}} $$

I tried dividing both sides by $A$ and then multiplying by $(j-x)$, but I'm not sure how to proceed from there because the $j$ is throwing me off. Usually it's a '$1$'.

Calvin Khor
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BirbCS
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1 Answers1

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The solutions of the equation (not depending on $A$, because its value can be any), are: $j=1$,

$j=\frac{x(x^{n}-x)}{1-x} $.

Verification for $j=$1, we take it for granted.

Verification for $j=\frac{x(x^{n}-x)}{1-x} $:

$A(j+x+x^2+....+x^n)= A(j-x{^{n+1}})(j-x){^{-1}}$,

$A(\frac{x(x^{n}-x)}{1-x}+ x+x^2+....+x^n)=\frac{A(\frac{x(x^{n}-x)}{1-x}-x^{n+1})}{\frac{x(x^{n}-x)}{1-x}-x}$,

$A(\frac{x(x^{n}-x)}{1-x}+ \frac{x(x^{n}-1)}{x-1})=\frac{A(\frac{x^{2}(x^{n}-1)}{1-x})}{ -\frac{x(x^{n}-x)}{1-x}+x}$,

$Ax=Ax$.

The assertion is true if

$j=1$,

and

$j=\frac{x(x^{n}-x)}{1-x} $.

giuseppe mancò
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