One part of a problem requires me to find following sum $\ x+x^2+x^3+...+x^n\ $ and solution suggests that after first step given sum equals to $ \left(x \frac{1-x^n}{1-x} \right) $ and I don't see how to get that. Could anyone explain it to me?
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$\begingroup$ FYI, the sum is of a geometric series. $\endgroup$– John OmielanCommented Apr 30, 2020 at 20:56
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$\begingroup$ I'm embarrassed. I constantly keep forgetting that formula for sum of elements of geometric series exist... It even costed me lots of points on finals recently :-( $\endgroup$– NigruteenCommented Apr 30, 2020 at 21:10
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2 Answers
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Let $$\alpha:=x+\cdots+x^n.$$ Then $$\alpha x=x^2+\cdots+x^{n+1}.$$ Therefore, $$\alpha(x-1)=x^{n+1}-x=x(x^n-1).$$ If $x\neq 1$, divide both sides with $x-1$, so $$\alpha=x\frac{x^n-1}{x-1},$$ the result you want. If $x=1$, $\alpha=n$.
answered Apr 30, 2020 at 20:55
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Multiply $x + x^2 + \cdots + x^n$ by $1-x$ and rearrange terms, you get $$\begin{array}{c} x &+& \color{red}{x^2} &+& \color{green}{x^3} &+& \cdots &+&\color{blue}{x^n}\\ &-& \color{red}{x^2} &-& \color{green}{x^3} &-& \cdots &-&\color{blue}{x^n} &-& x^{n+1} \end{array}$$ Notice the massive cancellation of terms, the result simplifies to
$$(1-x)(x + x^2 + \cdots + x^n) = x - x^{n+1} = x(1-x^n)$$ Divide both sides by $1-x$, you get what you want to show.
answered Apr 30, 2020 at 21:01