<s>[INST] Sum of the first integer powers of $n$ up to k Pascal's triangle has a lot of interesting patterns in it; one of which is the triangular numbers and their extensions. Mathematically:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}$$
At first, we could guess that the next summation is:
$$\sum_{n=1}^kn^3 ?=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}$$
Yet this is off. However, it is off geometrically. Notice:
$$\left(\sum_{n=1}^kn^3\right)-\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}=error$$
$$k=1, r=0$$
$$k=2, r=0.25$$
$$k=3, r=1$$
$$k=4, r=2.5$$
$$k=5, r=5$$
$$k=6, r=8.75$$
...
Consider the ratios of the errors:
$$er(k)=\frac{r(k+1)}{r(k)}$$
$$k=1, r=udf$$
$$k=2, r=4$$
$$k=3, r=2.5$$
$$k=4, r=2$$
$$k=5, r=1.75$$
$$k=6, r=1.6$$
Then, rewriting the error as a function of n starting at k = 5:
$$1.75=2.5-\frac{1.5}{2}$$
$$1.6=2.5-\frac{1.5}{2}-\frac{1.5}{10}$$
$$1.5=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}$$
$$1.42857=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}-\frac{1.5}{21}$$
The denominators in the series are from pascals triangle: (3rd columns, or dependent again on the triangular numbers)
Then the total formula for equating the two is:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\right)-\left(\sum_{n=1}^kn^3\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)=0$$
Super interesting!
At least, I thought it was interesting how this the error is related back to the previous power's formula. Am I missing something obvious? Any input is greatly appreciated. (I'm not smart, so in the likely event I missed something obvious try not to be too harsh)
Update:
For the next power (4), I found the formula with trial and error:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\cdot\frac{\left(4k+1\right)}{5}\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)+\frac{1}{12}\left(k-1\right)k\left(k+1\right)k$$
Any ideas on power (5) and so on? I'll continue to try and generalize it.
 [/INST] We can use telescoping sums to find formula for $$\sum_1^n k^p  $$ for $p\ge 1$ 
Note that $$ (k+1)^2 - k^2 =2k+1 \implies$$
$$ \sum_1^n \big[(k+1)^2 - k^2\big]=\sum_1^n (2k+1)\implies $$
$$ (n+1)^2 -1=2 \sum_1^n k  + n\implies $$
$$\sum_1^n k = \frac {n(n+1)}{2}$$
Similarly we can find formula for $$\sum_1^n k^2   $$
by using $$(k+1)^3 - k^3 =3k^2+ 3k+1$$
and so forth.  
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