When counting FLOPS, one must often compute certain sums.
Prove the exact formula
$$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$
I have used the approximation $\sum_{k=1}^{n}f(k)\approx\int_{1}^{n}f(k)dk$ to find that
$$\sum_{k=1}^nk^3= \frac{n^4}{4} + LOT$$ Lot stands for lower order terms
how do I go from here to calculate the exact number of flops for $$\sum_{k=1}^nk^3$$
First note that
$$ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{[n^2 + 4(n+1)](n+1)^2}{4} = \frac{(n+2)^2(n+1)^2}{4} $$
Then proceed by induction.
If you are not a fan of induction, you can solve problems like these using something that's similar to the derivative for sums. If you have seen derivatives before, you may know the power rule, which shows that $\frac{d}{dx} x^n = \lim_ {h \to 0}\frac{(x+h)^n-x^n}{h} = nx^{n-1}$. The important aspect of this is that the highest power decreases by 1 each time we apply the derivative. The same thing happens with sums. In particular $$\frac{(x+1)^n-x^n}{1} = \sum_{k=0}^{n-1}{n\choose k}x^k = nx^{n-1}+LOT\ $$
If I take the 'derivative' of the sum enough times, the LOT will go away.
Here is how to work out this method for your sum
Derivative 1 $$\sum_{k=0}^{n}k^3 - \sum_{k=0}^{n-1}k^3 = n^3$$ Derivative 2 $$n^3-(n-1)^3=3n^2 + LOT$$ Derivative 3 $$3n^2 + LOT - (3(n-1)^2 + LOT) = 6n+LOT$$ Derivative 4 $$6n+LOT-(6(n-1)+LOT) = 6$$ Since we know that the 4th derivative is 6, then the first term in our polynomial is $\frac{6}{4!}$ (we divide by 4! to counteract the fact that each derivative multiplies by n, so we have multiplied by 432*1). So, we get rid of the 4th power term and apply this process again.
Derivative 1 $$\sum_{k=0}^{n}k^3-\left(\frac{n^4}{4}\right) - \left(\sum_{k=0}^{n-1}k^3- \left(\frac{(n-1)^4}{4}\right)\right) = n^3 - \frac{n^4}{4} + \frac{(n-1)^4}{4} = \frac{6n^2}{4} + LOT$$ From here, we can immediately skip to the 3rd derivative by ignoring the LOT, and get that the 3rd derivative is 3. So, we now have $$\sum_{k=0}^n k^3 = \frac{n^4}{4} + \frac{n^3}{2} +LOT$$ Applying this one final time would yield $$\sum_{k=0}^n k^3 = \frac{n^4}{4} + \frac{n^3}{2} +\frac{n^2}{4}$$
Obviously, this process takes a lot more work than induction, and if you already know what formula the sum should turn into, use induction instead. I only include this method because this is something that is quite straightforward to code, and so if you do decide to code this you can get a solution to any FLOPS question which asks you to simplify a sum.
