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I'm doing homework for a programming class and came across this problem. There's no directions besides what I've shown, so I don't even know what it's asking me to do. What makes the most sense for what it's asking me to solve?

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    $\begingroup$ I expect they are asking you to simplify those forms. The first one, for example, is $x^{1+\dots+N}=x^{\frac {n(n+1)}2}$. $\endgroup$ – lulu Feb 6 '16 at 0:27
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Hint

Use that $a^n\cdot a^m=a^{n+m}$ to simplify $a).$ You should obtain $x^{N(N+1)/2}.$

Use that $\log_a a^x=a^{\log_a x}=x$ to simplify $b),c),d).$ You should obtain $6,280$ and $N,$ respectively.

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  • $\begingroup$ If you have time could you elaborate on the steps taken to get part (a)? I used that identity to get $x^{1+2+...+N}$, but don't see where to go from there. $\endgroup$ – Austin Feb 6 '16 at 0:44
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    $\begingroup$ Your answer is correct. I don't know if you are expected to simplify it more. Anyway, I have used that the sum of the arithmetic progression $1,2,\dots, N$ is $N(N+1)/2.$ See en.wikipedia.org/wiki/Arithmetic_progression. $\endgroup$ – mfl Feb 6 '16 at 0:50
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I think what are you supposed to do is to sinplify them. For example $$x\cdot x^2=x^3$$ So you could do the same thing for the first expression, then go on with the others.

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  • $\begingroup$ But it's to the N, I think @lulu is right here if that's considered a more simplified version of (a), although (b) already seems simplified to me $\endgroup$ – Austin Feb 6 '16 at 0:34
  • $\begingroup$ actually nevermind I think (b) might just equal 6 $\endgroup$ – Austin Feb 6 '16 at 0:39
  • $\begingroup$ I was just pointing out that the question wants you to simplify, obviously thay for the first point you have $x^{1+2+\dotsc+N}$. Then you just need to simplify it more. But just giving you the answer wouldn't be the right way of helping you I think :) $\endgroup$ – Marius Feb 6 '16 at 0:40

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