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