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I am involved in a challenge, called the Press Up Challenge. This involves starting on day 1, for me the 1st January, and doing 1 press up. This is followed on day 2 with 2 press ups.

As you can see, day 32, the 1st February will involve 32 press ups. This morning I managed my 49 press ups but it is getting harder!

So far I am simply keeping a log of what I have completed by way of adding each days number to my running total.

Is there a mathematical equation that will allow me to work out how many press ups I have achieved and will have achieved on any given day?

grg
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Barrie Fitzpatrick
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2 Answers2

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Call the value $f(n)$. Then:

$$\begin{align}2f(n)&=(1+2+\cdots+n)+(n+n-1+\cdots+1)\\&=(1+n)+(2+n-1)+\cdots+(n+1)\\&=(n+1)+(n+1)+\cdots+(n+1)\quad(\text{$n$ times})\\&=n(n+1)\end{align}$$

So: $$f(n)=\frac{1}{2}n(n+1)$$

grg
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Uncountable
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  • Many thanks for the immediate and very helpful response. I have used it to check back on my existing records and it works perfectly, I can only assume it will work for future numbers too. I will take some time to look at it and see if I can work out why it works and how you came to it, but in the mean time, Thank you once again. – Barrie Fitzpatrick Feb 18 '15 at 17:41
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    Of course it will work for future numbers, that's the power of mathematics :) Good luck with your press-ups. – Uncountable Feb 18 '15 at 18:19
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Another way: write the sum as $$S= 1+ 2+ 3+ ...+ (n-1)+ n$$ and then as $$S= n+(n-1)+ (n-2)+ ...+ 2+ 1$$

Adding vertically that gives \begin{align*}2S&= (n+1)+ (2+ n-1)+ (3+ n-2)+ ...+ (n-1+ 2)+ (n+ 1) \\ &= (n+1)+ (n+1+ (n+ 1)+ ...+ (n+1)+ (n+ 1)\end{align*} And there are $n$ such sums so $$2S= n(n+ 1), S= \frac{n(n+1)}{2}.$$

(That method is attributed to Gauss when he was a child.)

Hirshy
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user247327
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  • Please start using LaTeX, see this guide: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference Almost all of your posts need heavy editing – Hirshy Aug 05 '15 at 08:37