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$$p^{n+1} = p^0+p^1+ \dots + p^n$$


$$p^{n+1} = p^0\times p^1\times \dots \times p^n\text{ ?}$$

I am confused.

please explain the correct one.

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They are both wrong, as you will see if you take, say, $p=2$ and $n=3$. –  Gerry Myerson May 24 '13 at 11:56

3 Answers 3

up vote 1 down vote accepted


$$\begin{align*}\bullet&\;\;\;\;1+q+q^2+\ldots+q^n=\frac{q^{n+1}-1}{q-1}\\\bullet\bullet&\;\;\;\;1+2+\ldots+n=\frac{n(n+1)}2\\\bullet\bullet\bullet&\;\;\;a\cdot a^2\cdot\ldots\cdot a^n=a^{1+2+\ldots+n}\end{align*}$$

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Am i correct that $p^{n+1} = p \times p^n = p × \dots \times p$ [(n+1) times]. –  tree May 24 '13 at 12:17
Yes, you are, since $\,p^n=p\cdotp\cdot\ldots\cdot p\,\,\,( n\,$ times ) –  DonAntonio May 24 '13 at 12:32

None of them. $p^{n+1} = p \times p^n = p × \dots \times p$ (n times).

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is it n times or (n+1) times ? –  tree May 24 '13 at 12:09
@tree It's n times for multiplication, and (n+1) instances of the p symbol. –  Doug Spoonwood May 24 '13 at 15:51


$$a^m\cdot a^n=a^{m+n}$$

and $$\sum_{0\le m\le n}ar^m=\frac{a(r^{m+1}-1)}{r-1}$$

Now, judge yourself

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