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I was reading the proof for the sum of geometric progression at http://www.proofwiki.org/wiki/Sum_of_Geometric_Progression

and one of the statements is the following:

$$\sum_{j=1}^{n-1}{x^j}-\sum_{j=0}^{n-1}{x^j}=x^n+\sum_{j=1}^{n-1}{x^j}-(x^0+\sum_{j=1}^{n-1}{x^j})$$

I tried to decipher why this is true but I failed. How is the above statement derived?

Alby
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1 Answers1

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I assume you're referring to Proof 2, in which case you've copied the equality incorrectly; it should read:

$$\sum_{j=1}^{n}{x^j}-\sum_{j=0}^{n-1}{x^j}=x^n+\sum_{j=1}^{n-1}{x^j}-(x^0+\sum_{j=1}^{n-1}{x^j})$$

Further, notice that $$\sum_{j=0}^{n-1} x^j = x^0+\sum_{j=1}^{n-1}x^j.$$

And,

$$\sum_{j=1}^{n}x^j=x^n+\sum_{j=1}^{n-1}x^j.$$

Thus, $$\sum_{j=1}^{n}{x^j}-\sum_{j=0}^{n-1}{x^j}=x^n+\sum_{j=1}^{n-1}x^j-\left(x^0 + \sum_{j=1}^{n-1}x^j\right).$$

Your confusion may be coming from the following:

$$x\sum_{j=0}^{n-1}{x^j}=\sum_{j=0}^{n-1}x\cdot{x^j}=\sum_{j=0}^{n-1}{x^{j+1}}=\sum_{j=1}^{n}{x^j},$$

where in the last step we let $j \mapsto j-1$ and thus needed to shift the indices from $0, \dots, n-1$ to $1, \dots, n$.

doppz
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  • Just corrected the mistake! My confusion comes from $\sum_{j=0}^{n-1}{x^j}=x^0+\sum_{j=1}^{n-1}{x^j}$...I still don't see why this is the case – Alby Dec 07 '13 at 21:33
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    Ah, well in that case it's best to write it out. $\sum_{j=0}^{n-1}x^j=x^0+x^1+x^2+\dots+x^{n-1}=x^0+(x^1+x^2+\dots+x^{n-1})=x^0+ \sum_{j=1}^{n-1}x^j$. – doppz Dec 07 '13 at 21:56