Sums of Powers

We now turn to the relation between the Bernoulli polynomials and h-calculus. By Proposition 23.1, we have $$ D_1 B_n (x) = B_n (x + 1) - B_n (x) = nx^{n - 1} , $$ or (24.1)$$ n\int {x^{n - 1} d_1 x = B_n (x),} $$ where D₁ is the h-derivative with h = 1 and ∝ f(x)d₁x stands for the h-antiderivative with h=1. Applying the fundamental theorem of h-calculus (22.14), we have for a nonnegative integer n, (24.2)$$ a^n + (a + 1)^n + \cdots + (b - 1)^n = \int_a^b {x^n d_1 x = \frac{{B_{n + 1} (b) - B_{n + 1} (a)}} {{n + 1}}} , $$ where a < band b - a ∈ ℤ. If we rewrite the right-hand side using (23.5) and let a = 0, b = M + 1, we get (24.3)$$ \sum\limits_{k = 0}^M {k^n } = \frac{1} {{n + 1}}\sum\limits_{j = 0}^n {\left( {_j^{n + 1} } \right)} (M + 1)^{n + 1 - j} b_j . $$