Polynomial Formula for Series So far, I'm stuck on this problem of converting a series to a polynomial and showing that it exhibits certain properties.
PROBLEM

Show that the polynomial formula for $P_k(n) = \sum_{j=1}^n j^k$ is characterized by the following two properties:
  
  
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*$P_k(0) = 0$ for all $k$
  
*$P_k(x)-P_k(x-1) = x^k$ for all $k$ and all $x$
  
  
  Then show that $P_{k+1}(x) = (k+1)\int_0^x P_k(t) \, dt + C_{k+1} x$ for some constant $C_{k+1}$.

By "polynomial formula", we mean that $P_1(n) = 1 + 2 + ... + n = \frac 12 n^2 + \frac 12 n$ for all positive integer $n$, so we extend $P_1(x) = \frac 12 x^2 + \frac 12 x$ for all real $x$.
MY APPROACH
I showed that any polynomial that satisfies those two properties is unique, so that, if that integral expression (call it $R_{k+1}(x)$) satisfies them with $R_{k+1}(x)-R_{k+1}(x-1)=x^{k+1}$, then we must have $P_{k+1} = R_{k+1}$.
I'm just not sure how to show that $R_{k+1}$ exhibits the second property. Here's what I have:
$R_{k+1}(x) - R_{k+1}(x-1) = (k+1)\int_{x-1}^x P_k(t) \, dt + C_{k+1}$
Expanding $P_k(t) = p_0 + p_1 t + p_2 t^2 + ...$ gives
$R_{k+1}(x) - R_{k+1}(x-1) = (k+1)\int_{x-1}^x \sum_{i=0}^\infty p_i t^i \, dt + C_{k+1}$
$ = (k+1)\sum_{i=0}^\infty \frac{p_i}{i+1}(x^{i+1}-(x-1)^{i+1}) + C_{k+1}$
Any advice on where to go from here? Thanks in advance.
 A: I got it! (I think.)
SOLUTION
Let $R_{k+1}(x) = (k+1)\int_0^x P_k(t) \,dt + C_{k+1}x$. Then note that $R_{k+1}(0) = (k+1)\int_0^0 P_k(t) \,dt + C_{k+1} \cdot 0 = 0$.
For the second property:
By hypothesis, we have $P_k(t) - P_k(t-1) = t^k$ 
Integrating from $0$ to $x$:
$\int_0^x P_k(t) \, dt - \int_0^x P_k(t-1) \, dt = \int_0^x t^k \, dt = \frac{1}{k+1}x^{k+1}$
Multiplying by $k+1$ and letting $u=t-1$ in the second integral:
$(k+1)\int_0^x P_k(t) \, dt - (k+1)\int_{-1}^{x-1} P_k(u) \, du = x^{k+1}$
Splitting the second integral at 0:
$(k+1)\int_0^x P_k(t) \, dt + \underset{C_{k+1}}{\underbrace{^- (k+1)\int_{-1}^{0} P_k(u) \, du}} - (k+1)\int_0^{x-1} P_k(u) \, du = x^{k+1}$
But $C_{k+1} = C_{k+1} \times \left[x - (x-1) \right]$, so
$(k+1)\int_0^x P_k(t) \, dt + C_{k+1}x - \left[(k+1)\int_0^{x-1} P_k(u) \, du + C_{k+1}(x-1)\right] = x^{k+1}$
But this is precisely
$R_{k+1}(x) - R_{k+1}(x-1) = x^{k+1}$
Hence, by the uniqueness of $P_{k+1}$, we must have $P_{k+1}(x) = (k+1)\int_0^x P_k(t) \,dt + C_{k+1}x$.
