Prove by Induction with Summation Struggling with this , especially with the double summation; if anyone can help it be much appreciated!
$$\forall n \in \Bbb N : \quad \sum_{i=1}^n \sum_{c=1}^i c = \sum_{i=1}^n i(n − i + 1) .$$
It needs to be answered in the following format:
1- Prove for n=1
2- assume n=k
3- prove for k+1

Here  is   an  answer  based upon induction. We show:

The following is valid
  \begin{align*}
\sum_{i=1}^n \sum_{c=1}^i c = \sum_{i=1}^n i(n - i + 1)\qquad\qquad n\geq 1
\end{align*}

Base step: $n=1$

We have to show
  \begin{align*}
\sum_{i=1}^1 \sum_{c=1}^i c = \sum_{i=1}^1 i(1 - i + 1)
\end{align*}
Since
  \begin{align*}
\text{LHS: }\qquad&\sum_{i=1}^1 \sum_{c=1}^i c =\sum_{c=1}^1 c=1\\
\text{RHS: }\qquad&\sum_{i=1}^1 i(1 - i + 1)=1\cdot(1-1+1)=1
\end{align*}
the claim is valid for $n=1$.

Induction hypothesis: $n=k$

We assume the validity of
  \begin{align*}
\sum_{i=1}^k \sum_{c=1}^i c = \sum_{i=1}^k i(k - i + 1)\tag{1}
\end{align*}

Induction step: $n=k+1$

We have to show
  \begin{align*}
\sum_{i=1}^{k+1} \sum_{c=1}^i c = \sum_{i=1}^{k+1} i(k - i + 2)
\end{align*}
We obtain
  \begin{align*}
\sum_{i=1}^{k+1} \sum_{c=1}^i c&=\sum_{i=1}^{k} \sum_{c=1}^i c+\sum_{c=1}^{k+1} c\tag{2}\\
&=\sum_{i=1}^{k} i(k - i + 1)+\sum_{c=1}^{k+1} c\tag{3}\\
&=\sum_{i=1}^{k+1} i(k - i + 1)+\sum_{c=1}^{k+1} c\tag{4}\\
&=\sum_{i=1}^{k+1} i(k - i + 1)+\sum_{i=1}^{k+1} i\tag{5}\\
&=\sum_{i=1}^{k+1} i(k - i + 2)\tag{6}\\
\end{align*}
  and the claim follows.

Comment:


*

*In (2) we separate the summand with index $i=k+1$.

*In (3) we apply the induction hypothesis (1).

*In (4) we set the upper limit of the left sum to $k+1$ without changing anything since we are adding zero only.

*In (5) we change the name of the index $c$ to $i$ in the second sum.

*In (6) we merge the second sum into the first one.
