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
 A: If you make a drawing in the plane of the couples $(i,c)$ that are being considered in the LHS sum, you'll see a triangle with apexes $(1,1)$, $(n,1)$ and $(n,n)$. In this triangle, you count the $c$'s vertically.
Here is a visualization of what is happening:

Now switch the two sums, using your drawing :
$$\sum_{i=1}^n \sum_{c=1}^i c = \sum_{c=1}^n \sum_{i=c}^n c$$
but in the inner RHS sum, $c$ is a constant, and there are $n-c+1$ terms. So :
$$\sum_{c=1}^n \sum_{i=c}^n c = \sum_{c=1}^n c(n-c+1)$$
All that is left is to rename $c$ as $i$ :-)
I'd like to include the drawing, but don't know how to do. Maybe Tikz is implemented ?
A: $$\begin{align}
\sum_{i=1}^n \sum_{c=1}^i c 
&= \sum_{c=1}^n \sum_{i=c}^n c\qquad (1\le c\le i\le n)\\
&= \sum_{c=1}^n c\sum_{i=c}^n 1\\
&=\sum_{c=1}^n c(n-c+1)\\
&=\sum_{i=1}^n i(n − i + 1)
\end{align}$$
A: Rewrite the identity for $n+1$:


*

*Left hand side
$$\sum_{i=1}^{n+1} \sum_{c=1}^i c =\sum_{i=1}^{n} \sum_{c=1}^i c+\sum_{c=1}^{n+1} c.$$

*Right hand side
$$\sum_{i=1}^{n+1} i(n+1 − i + 1)=\sum_{i=1}^{n+1} i(n− i + 1)+\sum_{i=1}^{n+1}i=\sum_{i=1}^{n} i(n− i + 1)+\sum_{i=1}^{n+1}i.$$
As the additional terms are the same, if the identity holds for $n$, then it holds for $n+1$.
A: 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.
