Problem source

The \textit{Bernoulli polynomials} $P_{n}(x)$, where $n$ is a non-negative integer, are defined by $P_{0}(x)=1$ and, for $n\geqslant1$, 
\[
\frac{\mathrm{d}P_{n}}{\mathrm{d}x}=nP_{n-1}(x),\qquad\int_{0}^{1}P_{n}(x)\,\mathrm{d}x=0
\]
Show by induction or otherwise, that 
\[
P_{n}(x+1)-P_{n}(x)=nx^{n-1},\quad\mbox{ for }n\geqslant1.
\]
Deduce that
\[
n\sum_{m=0}^{k}m^{n-1}=P_{n}(k+1)-P_{n}(0)
\]
Hence show that ${\displaystyle \sum_{m=0}^{1000}m^{3}=(500500)^{2}}$

Solution source

$\displaystyle \int_x^{x+1} nP_{n-1}(x) \, dx = P_n(x+1) - P_n(x)$

Claim: $P_{n}(x+1)-P_{n}(x)=nx^{n-1},$ for $n \geq 1$

Proof: (By induction). (Base case, $n=1$).

$P_1(x) = x - \frac12$, $P_1(x+1) - P_1(x) = 1 x^{0}$ as required.

Assume the equation is true for $n = k$. So $P_k(x+1) - P_k(x) = kx^{k-1}$ now consider 
\begin{align*}
P_{k+1}(x+1) - P_{k+1}(x) &= \int_0^{x+1} (k+1) P_k(t) \d t + P_{k+1}(0)-  \int_0^{x} (k+1) P_k(t) \d t - P_{k+1}(0) \\
&= \int_0^x (k+1)(P_k(t+1)-P_k(t)) \d t + \int_0^1 (k+1)P_k(t) \d t \\
&= (k+1)x^{k} + 0
\end{align*}
So by induction we are done.

\begin{align*}
n\sum_{m=0}^{k}m^{n-1} &= \sum_{m=0}^{k}n \cdot m^{n-1} \\
&=  \sum_{m=0}^{k}\l P_n(m+1)-P_n(m) \r \\
&= P_n(k+1) - P_n(0)
\end{align*}

We need to find $P_4$


\begin{align*}
P_0(x) &= 1 \\
P_1(x) &= x - \frac12 \\
P_2(x) &= x^2 -x - \int_0^1 \l x^2 - x \r \d x \\
&= x^2 - x + \frac16 \\
P_3(x) &= x^3 -\frac{3}{2}x^2 + \frac12x -  \int_0^1 \l x^3 -\frac{3}{2}x^2 + \frac12x \r \d x \\
&= x^3 -\frac{3}{2}x^2 + \frac12x \\
P_4(x) &= x^4 - 2x^3 + x^2 + c
\end{align*}

Therefore the sum we are interested in is $\frac14 \l P_4(1001) - P_4(0) \r =  \frac14 (1001)^2 (1001-1)^2 = (1001 \cdot 500)^2 = (500500)^2$