Why 1+2 + … + n = n × (n+1) /2 ?

Still remember how to calculate the summation for 1+2+3+ … + 10 ? Well, the summation for the first ten sequence (starting from 1) is 55.

But, how do you calculate it? Pressing the sequence into calculator, one by one manually?

Fine. What if I change the question to 1+2+ …+ 1000. Probably, you will not bother to calculate this anymore.

Try to recall what you have learned in secondary school  — there is a formula!

1+2+3+ …+ n = n × (n + 1) /2

Perhaps, the math teacher had challenged you to add from 1 to 100, or to any suggested number and he could give you the answer in 5 seconds. Later, he disclosed the secret formula to you and the whole class laughing together and learned the secret formula.

So, why 1+2+3+ …+ n = n × (n + 1) /2  ?

Perhaps your math teacher just ask you memorize this secret formula (that why you have forgotten?). In fact, there are a few proofs to derive this formula. In the following, I will provide a simple example to derive the summation formula. Be noted that, this is just a simple illustration and it is NOT a complete proof.

Okay. Assuming we have 4 x 4 = 16 balls that are arranged as the following picture.

This balls topology can further be segmented to two triangles one rectangle.

So, the first triangle + the middle rectangle +right triangle balls = 16 balls. Note that we didn’t change the number of balls. We just represent 16 balls in different topology.
The left and the triangles are actually the same balls. So, they can be combined as “two triangles”.
Okay. We continue to move the second rectangle to the right. So, the right hand side becomes 4×4 – the rectangle.
Since the rectangle has 4 balls, 4 x 4 – rectangle becomes 4 x 3 (or, 3 x 4).
Now, we divide the left hand and the right hand sides by 2, and the “equation” becomes “one triangle = 4 x 3 /2”.
Note that the triangle is one ball + two balls + three balls. If we change all these balls to numbers, the equation becomes…
That is exactly 1 + 2 + … + n = n × (n + 1) / 2.
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The example is taken from Joseph H. Silverman’s “A Friendly Introduction to Number Theory” 3rd Edition.