A mathematician Johann Carl Friedrich Gauss, born in 1777 in Germany is one of my heroes in mathematics. In his earlier years he was a disruptive child. He would distract children in class and not listen to what the teacher taught him as he believed he already knew everything that he could teach him.

One day, his teacher set the class a punishment for their bad behaviour. He said "Calculate the sum of all the numbers from one to one hundred". The teacher than sat at their desk hoping that it would keep them quiet, but a short while in, Gauss had completed it. His teacher, in outrage and disbelief charged over to him and saw that he had written the number 5050 on his page.

The teacher didn't believe him and thought he had cheated. "Ok, well, find the sum up to 200 then" they said, and before the teacher had even reached his desk Gauss shouted again. Confused, the teacher walked back to him and demanded to know how he did it so fast. He had used the following method:

He knew that he had to do 1+2+3+4...+99+100. so he thought, why not write it out backwards. so he wrote it like this:

\(1+2+3\quad ...\quad +98+99+100\)

\(100+99+98\quad ...\quad +3+2+1\)

\(101+101+101\quad ...\quad +101+101+101\)

\({ S }_{ n }=\frac { 1 }{ 2 } ({ n(n+1) }\quad =>\quad { S }_{ n }=\frac { (100(101)) }{ 2 } =\quad 5050\)

It was this that then lead onto the summation of an arithmetic sequence. An equation which is still in use today:

\( S_n=\frac { n }{ 2 } ({ 2a }_{ 1 }+(n-1)d) \)

A small realization that you can use a variant on this to find the sum of times tables

with the 1 times table:

1, 2, 3, 4, 5, 6, 7, 8, 9 -> The sum progresses as 1, 3, 5, 10, 15, 21.... This is a formula of \( (n^2+n) /2 \).

And with the two times table:

2, 4, 6, 8, 10, 12, 14 -> This sum progresses as 2, 6, 12, 20, 30... This is a formula of \((n^2 +n)\) which is twice the formula of the equation for the 1x table.

It turns out that this pattern will continue in that the sum will be:

\(S_n=\frac { 1 }{ 2 } { (n }^{ 2 }+n)\quad +\quad { (n }^{ 2 }+n)\quad +\quad \frac { 1 }{ 2 } ({ 3n }^{ 2 }+3n) \)

and then multiplying by 2 will give:

\(2S_n={ (n }^{ 2 }+n)\quad +\quad { (2n }^{ 2 }+2n)\quad +\quad ({ 3n }^{ 2 }+3n) \)

This will then calculate the sum of any times table up to the nth term in each but the following equation is the final equation of this paper and is the sum to infinity of the above equation written in terms of n and x where x is the times table and n is the term in that times table:

\( { S }{ \infty }=\frac { 1 }{ 2 } (\sum _{ x,n=1 }^{ { n(x) }{ \infty } }{ x({ n }^{ 2 } } +n)) \)

## Comments

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TopNewestThanks for sharing! Note that to get the math to display, you have to place it with the brackets. I've edited your post slightly to give you an idea. – Calvin Lin Staff · 2 years, 1 month ago

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– Jack Barker · 2 years, 1 month ago

Thank you! i was having difficulty with thisLog in to reply

Nice note but wrap maths terms properly – Pranjal Jain · 2 years, 1 month ago

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