# Why doesn't this formula work in summation of cubes? but works in summation of squares

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

right?

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

• How does it not work? – Karolis Juodelė Nov 12 '12 at 15:34
• Do you understand that $k$ is not a constant here, so it has no meaning outside the $\sum$? – Thomas Andrews Nov 12 '12 at 15:34
• I'm not quite sure what your question is here. What exactly are you trying to show? – Michael Dyrud Nov 12 '12 at 15:34
• 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.) – Thomas Andrews Nov 12 '12 at 15:48
• 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." – Thomas Andrews Nov 12 '12 at 15:52

Ignoring your title, if you are asking for $$\sum^{n}_{k=j} k^3$$ (so starting at $j$ instead of $1$) you can write $$\sum^{n}_{k=j} k^3 = \sum^{n}_{k=1} k^3 -\sum^{j-1}_{k=1} k^3=({n^2(n+1)^2})/4-(j^2(j-1)^2)/4$$
I'll try to answer the question as best I can. For any sum, the following identity holds, where $n > m > a$:
$$\sum_{k=a}^{n}{f(k)}=\sum_{k=a}^{m}{f(k)}+\sum_{k=m+1}^{n}{f(k)}$$
$$\sum_{k=a}^{n}{f(k - b)}=\sum_{k=a-b}^{n-b}{f(k)}$$ So: $$\sum_{k=b}^{n}{f(k)}=\sum_{k=a}^{n}{f(k)}-\sum_{k=a}^{b-1}{f(k)}$$