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$$ \sum^{n}_{k=1} k^3 = ({n^2(n+1)^2})/4 $$

right?

say for example k not equal to 1, why doesn't this work? I subtracted the summation of k-1? $$ \sum^{n}_{k!=1} k^3 = ({n^2(n+1)^2 - (k-1)^2k^2})/4 $$

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    How does it not work?2012-11-12
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    Do you understand that $k$ is not a constant here, so it has no meaning outside the $\sum$?2012-11-12
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    I'm not quite sure what your question is here. What exactly are you trying to show?2012-11-12
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    say for example k starts at a number other than 1, like 32012-11-12
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    Presumably, he means $$\sum_{k\neq 1}^n$$, which is meaningless2012-11-12
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    @ThomasAndrews meaningless how?2012-11-12
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    So you mean $$\sum_{k=j}^n k^3 = \frac{{n^2(n+1)^2 - (j-1)^2j^2}}4$$2012-11-12
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    @ThomasAndrews yeah exactly!2012-11-12
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    @ThomasAndrews because I got the formula by subtracting the summation of j, I thought it would work but then I checked my answer using a calculator. So how do you derive it if it starts with j?2012-11-12
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    Do you have an example $n,j$ for which it does not work? Works for me.2012-11-12
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    @ThomasAndrews nevermind, double checked it again. Although I thought you said it was meaningless?2012-11-12
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    Your notation was meaningless - I had to ask several questions to figure out what you meant by your notation. $$\sum_{k\neq 1}^n$$ is not a valid mathematical formula (and there was no interpreation of it that gave me what you meanted to ask, since you meant $k=j$ or $k>j-1$ or some such.)2012-11-12
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    Also, there is no way for $\sum_k$ to end up being a function of $k$ because $\sum_k$ is an expression that means as $k$ varies, so the right side can't have $k$ in it - there is no one value of $k$ to put on the right side. That was why I added the variable $j$ as the "starting point."2012-11-12

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Ignoring your title, if you are asking for $$\sum^{n}_{k=j} k^3 $$ (so starting at $j$ instead of $1$) you can write $$\sum^{n}_{k=j} k^3 = \sum^{n}_{k=1} k^3 -\sum^{j-1}_{k=1} k^3=({n^2(n+1)^2})/4-(j^2(j-1)^2)/4$$

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    Beat me to it :)2012-11-12
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I'll try to answer the question as best I can. For any sum, the following identity holds, where $n > m > a$:

$$\sum_{k=a}^{n}{f(k)}=\sum_{k=a}^{m}{f(k)}+\sum_{k=m+1}^{n}{f(k)}$$

From this, we can note and conclude, via substitutions:

$$\sum_{k=a}^{n}{f(k - b)}=\sum_{k=a-b}^{n-b}{f(k)}$$ So: $$\sum_{k=b}^{n}{f(k)}=\sum_{k=a}^{n}{f(k)}-\sum_{k=a}^{b-1}{f(k)}$$