Prove by induction $\sum\limits_{k=0}^{n}( k+1)( n-k+1) =\binom{n+3}3$ Prove by induction :
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\binom{n+3}{3}$$
induction basis: $n=0$
$$\displaystyle \sum _{k=0}^{n}( 0+1)( 0-0+1) =1=\binom{3}{3}$$
For $n+1$:
\begin{aligned}
\sum _{k=0}^{n+1}( n+1+1)( n+1-(n+1)+1) & =\sum _{k=0}^{n}( k+1)( n-k+1) +( n+2)(( n+1) -( n+1) +1)\\
 & =\binom{n+3}{3} +( n+2)\\
\end{aligned}
I had to stop here because I realize there is a mistake ...
Unfortunately, I didn't succeed in many ways.

You set up the induction correctly by getting the base case and stating your induction hypothesis.
For the induction step:
$\sum\limits_{k=0}^{n+1}(k+1)((n\color{red}{+1})-k+1)$  You seem to have forgotten this red $\color{red}{+1}$.
So, we have $\sum\limits_{k=0}^{n+1}(k+1)((n\color{red}{+1})-k+1) = \sum\limits_{k=0}^n(k+1)(n-k+1)\color{red}{+\sum\limits_{k=0}^n(k+1)}+(n+2)$.
This should hopefully get you back on track.

 The first summation simplifies to $\binom{n+3}{3}$ per induction hypothesis, the second summation simplifies to $\binom{n+2}{2}$ recognizing it as the $n+1$'st triangular number.  So we have $\binom{n+3}{3}+\binom{n+2}{2}+\binom{n+2}{1}=\binom{n+3}{3}+\binom{n+3}{2}=\binom{n+4}{3}=\binom{(n+1)+3}{3}$ using Paschal's identity to finish.

