Suppose that $x\ne 1$ and $n\in\mathbb{N}^*$. Prove with induction that $$\sum_{k=0}^{n-1}x^{k}=\frac{x^n-1}{x-1}$$
It seems simple but I have tried for I don't know how long by now... Anyone can manage this?
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4Hint: $$\sum_{k=0}^n x^k=x^n+\sum_{k=0}^{n-1} x^k=x^n+\frac{x^n-1}{x-1}=\frac{x^n(x-1)+x^n-1}{x-1}=\frac{x^{n+1}-1}{x-1}$$ – learner May 13 '16 at 12:48
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4@learner it's not a hint, it's the whole answer. – Dark May 13 '16 at 13:00
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1@Dark The comment of learner is not the whole induction procedure. – callculus42 May 13 '16 at 13:10
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It is also easy to see without using induction just by moving the denominator to the left side and noting that everything cancels. – Vik78 May 13 '16 at 14:35
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Let $S(n)$ be the statement: $\displaystyle\sum_{k=0}^{n-1}x^{k}=\dfrac{x^{n}-1}{x-1}$
Basis step: $S(1)$:
LHS: $\displaystyle\sum_{k=0}^{(1)-1}x^{k}=x^{0}$
$\hspace{23 mm}=1$
RHS: $\dfrac{x^{(1)}-1}{x-1}=\dfrac{x-1}{x-1}$
$\hspace{27 mm}=1$
$\hspace{73.5 mm}$ LHS $=$ RHS $\hspace{1 mm}$ (verified.)
Inductive step:
Assume $S(m)$ is true, i.e. assume that $\displaystyle\sum_{k=0}^{m-1}x^{k}=\dfrac{x^{m}-1}{x-1}$
$S(m+1)$: $\displaystyle\sum_{k=0}^{(m+1)-1}x^{k}$
$\hspace{17 mm}=\displaystyle\sum_{k=0}^{m}x^{k}$
$\hspace{17 mm}=x^{m}+\displaystyle\sum_{k=0}^{m-1}x^{k}$
$\hspace{17 mm}=x^{m}+\dfrac{x^{m}-1}{x-1}$
$\hspace{17 mm}=\dfrac{x^{m}\hspace{1 mm}(x-1)+x^{m}-1}{x-1}$
$\hspace{17 mm}=\dfrac{x^{m+1}-x^{m}+x^{m}-1}{x-1}$
$\hspace{17 mm}=\dfrac{x^{m+1}-1}{x-1}$
So, $S(m+1)$ is true whenever $S(m)$ is true.
Therefore, $\displaystyle\sum_{k=0}^{n-1}x^{k}=\dfrac{x^{n}-1}{x-1}$.
Tazwar
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We give some hints for your question:
1) $S_{n+1}-S_n=x^n$
2) $\displaystyle\frac{x^{n+1}-1}{x-1}-\displaystyle\frac{x^{n}-1}{x-1}=x^n$.
Blind
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