Count To A & Count To B

Relevant wiki: Sum of n, n², or n³

It's much simpler to explain to show that

$(1+2+3+\cdots+ A) + (1+2+3+\cdots + B) + (A \times B )= 1 + 2 + 3 + \cdots + (A+ B)$

is true with a visual aid.

For simplicity sake, I let $A = 6$ and $B = 7$ .

The total number of red dots is $A \times B$ .
The total number of green dots is $1 + 2 + 3 + \cdots + A$ .
The total number of blue dots is $1 + 2 + 3 + \cdots + B$ .

So, the total number of dots is $(1+2+3+\cdots+ A) + (1+2+3+\cdots + B) + (A \times B )$ .

By counting the total number of dots in the overall picture, we have $1 + 2 + 3 + \cdots + (A+B)$ .

Hence, $(1+2+3+\cdots+ A) + (1+2+3+\cdots + B) + (A \times B )= 1 + 2 + 3 + \cdots + (A+ B)$ is true!

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Alternatively, here's an algebraic proof:

We apply the algebraic identity $1 + 2 + 3 + \cdots + n = \dfrac12 n(n+1)$ .

We have
$(1+2+3+\cdots+ A) = \dfrac12 A(A+1)$ ,
$(1+2+3+\cdots + B) = \dfrac12B(B+1)$ , and
$1 + 2 + 3 + \cdots + (A+ B) = \dfrac12 (A+B)(A+B+1)$ .
So,

$\begin{aligned} (1+2+3+\cdots+ A) + (1+2+3+\cdots + B) + (A \times B ) &=& \dfrac12 A(A+1) + \dfrac12B(B+1) + AB \\ &=& \dfrac12 ( A^2 + A + B^2 + B + 2AB) \\ &=& \dfrac12 (A+B)(A+B+1) \\ &=& 1 + 2 + 3 + \cdots + (A+ B) \\ \end{aligned}$

$\begin{aligned} S & = \color{#3D99F6}{(1+2+3+...+A)} + \color{#D61F06}{(1+2+3+...+B)} + AB \\ & = \color{#3D99F6}{\frac {A(A+1)}2} + \color{#D61F06}{\frac {B(B+1)}2} + AB \\ & = \frac {A^2}2 + \frac A2 + \frac {B^2}2 + \frac B2 + AB \\ & = \frac 12 \left(A^2 + 2AB + B^2\right) + \frac 12 (A+B) \\ & = \frac 12 \left(A+B\right)^2 + \frac 12 (A+B) \\ & = \frac {(\color{#3D99F6}{A+B})(\color{#3D99F6}{A+B}+1)}2 \\ & = \boxed{1 + 2 + 3 +...+ (\color{#3D99F6}{A+B})} \end{aligned}$