How can I find $\sum_{n=1}^N n^2-n$? Wolfram Alpha will tell you that it is $\frac{N}{3} (N-1)(N+1)$, and given the famous formulas for $\sum_{n=1}^N n^2$ and $\sum_{n=1}^N n$, you could piece together the first. But is there some sort of a general method here that might be of use in evaluating these kinds of partial sums?
Answer
I'm not sure what you mean by general but here are two ways to find $\sum_{n=1}^N p(n)$ for $p$ a polynomial.
If $p$ has degree $d$, find the value of the sum for $d+2$ values of $N$ and use Lagrange interpolation.
Write $p(n)$ in the binomial basis $n \choose k$ and use sum-of-column identity in the Pascal triangle.
No comments:
Post a Comment