Expand$\frac{ x^{k+1}}{k+1} -\frac{ (x-1)^{k+1}}{k+1} $using the binomial theorem The required is to expand 
$$\frac{x^{k+1}}{(k+1)} - \frac{(x-1)^{k+1}}{(k+1)}$$
using the binomial theorem.
Here is my solution, is it correct?
$$x^k -\frac{kx^{k-1}}{2} + \frac{\binom{k}{2}x^{k-2}}{3} - \frac{\binom{k}{3}x^{k-3}}{4} + \frac{\binom{k}{4}x^{k-4}}{5} + .... - \frac{\binom{k}{k - 3}x^{4}(-1)^{k-3}}{4} - \frac{\binom{k}{k - 2}x^{3}(-1)^{k-2}}{3} - \frac{kx^{2}(-1)^{k-1}}{2}  - x(-1)^{k} - \frac{-1^{k+1}}{k+1}$$
 A: The binomial theorem states:
$$
  \left(a + b\right)^n = \sum_{m=0}^n \binom{n}{m} \cdot a^m \cdot b^{n-m}
$$
Applying it for $(x-1)^{k+1}$, with $a = x$, $b=-1$ and $n=k+1$ we have:
$$\begin{eqnarray}
  \left(x-1\right)^{k+1} &=& \sum_{m=0}^{k+1} \binom{k+1}{m} \cdot x^m \cdot \left(-1\right)^{k+1-m} \cr &=& x^{k+1} + \sum_{m=0}^{k} \binom{k+1}{m} \cdot x^m \cdot \left(-1\right)^{k+1-m} \cr
  &=& x^{k+1} - \cdot \sum_{m=0}^{k} \binom{k+1}{m} \cdot x^{m} \cdot \left(-1\right)^{k-,}
\end{eqnarray}
$$
Now, observe that
$$
  \binom{k+1}{m} = \frac{\left(k+1\right)!}{\left(k+1-m\right)! \cdot m!} = \frac{k+1}{k+1-m} \cdot \frac{k!}{\left(k-m\right)! \cdot m!} = \frac{k+1}{k+1-m} \cdot \binom{k}{m}
$$
Combining these:
$$\begin{eqnarray}
 \frac{x^{k+1}-\left(x-1\right)^{k+1}}{k+1} &=& \cdot \sum_{m=0}^{k} \frac{1}{k+1-m} \binom{k}{m} \cdot x^{m} \cdot \left(-1\right)^{k-m} \cr
&\stackrel{m \to k - p }{=}& \cdot \sum_{p=0}^{k} \frac{1}{p+1} \binom{k}{k-p} \cdot x^{k-p} \cdot \left(-1\right)^{p} \cr &=& \cdot \sum_{p=0}^{k} \frac{1}{p+1} \binom{k}{p} \cdot x^{k-p} \cdot \left(-1\right)^{p}
\end{eqnarray}$$

Added:
Answering the OP's questions posed in comments, here is the verification of the result in Mathematica:

orig[k_Integer,x_]:=(x^(k+1)-(x-1)^(k+1))/(k+1)
expanded[k_Integer,x_]:=Sum[Binomial[k,p]/(p+1)(-1)^p x^(k-p),{p,0,k}]

In[3]:= Table[orig[k,x]==expanded[k,x]//Simplify,{k,0,6}]
Out[3]= {True,True,True,True,True,True,True}

