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The Bernoulli polynomials extend the Bernoulli numbers by inserting the exponential function $e^{xt}$ into their exponential generating function:For example, $B_0(x)=1$, $B_1(x)=x-1/2$, $B_2(x)=x^2-x+1/6$, and $B_3(x)=x^3-3x^2/2+x/2$.^[ref]

Their crucial property is a finite difference identity:Indeed, subtracting the two exponential generating functions cancels $e^t-1$ and leaves $t{e}^{xt}$. Thus $B_{p+1}(x)/(p+1)$ plays for discrete summation the role that $x^{p+1}/(p+1)$ plays for integration: taking a unit finite difference recovers $x^p$.^[ref]

An integer power sum adds a fixed nonnegative integer power of consecutive integers. Let $N$ be a nonnegative integer and $p$ a nonnegative integer. For $p=0$, use the constant polynomial $x^0=1$, including at zero, so $S_0(N)=N$. Summing the Bernoulli polynomial finite difference identity makes the intermediate terms cancel and givesThe sum is empty when $N=0$. For the endpoints $1,\dots ,N$, the formula valid also at $p=0$ is insteadFor $p\ge 1$, $B_{p+1}(1)=B_{p+1}(0)$, and this becomes the standard positive-endpoint formula. The exception at $p=0$ is essential: $B_1(1)-B_1(0)=1$.^[ref]

For $p\ge 1$, adding $N^p$ to $S_p(N)$ changes the $B_1$ contribution from $-N^p/2$ to $+N^p/2$. Equivalently,where only $B_1^{+}$ differs from our Bernoulli numbers. A quick endpoint check is $p=1$: the sums over $0,\dots ,N-1$ and $1,\dots ,N$ are $N(N-1)/2$ and $N(N+1)/2$, respectively. The latter is a triangular number.