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Joker
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Joined: 19 Mar 2011
Age: 37
Gender: Male
Posts: 7,593
Location: North Carolina The Tar Heel State :)
29 Apr 2012, 8:46 pm
I found this link it shows how you can turn math into a game it's very fun to play in your head.
Take a look at the link and tell me if you think the math games are fun or not 
http://www.angelfire.com/me/marmalade/mathtips.html
29 Apr 2012, 11:16 pm
Joker
Veteran
Joined: 19 Mar 2011
Age: 37
Gender: Male
Posts: 7,593
Location: North Carolina The Tar Heel State :)
30 Apr 2012, 12:35 am
30 Apr 2012, 3:00 am
01 May 2012, 6:53 am
Declension wrote:
It's maybe a slightly different definition of "trick", but one of my favourite "magic tricks" in maths is the "collapsing series" trick.
Here is an example: let's say that we are interested in what you get when you add up the first N powers of 2. We experiment by calculating the first few:
2^1 = 1.
2^1 + 2^2 = 2 + 4 = 6.
2^1 + 2^2 + 2^3 = 2 + 4 + 8 = 14.
2^1 + 2^2 + 2^3 + 2^4 = 2 + 4 + 8 + 16 = 30.
Well, is there any obvious pattern here? In other words, is there a nice formula which will quickly tell us what (2^1 + ... + 2^N) is, for any natural number N?
Here is the magic trick: we simply define S(N) = 2^1 + ... + 2^N, for any natural number N, and we say the following magic words:
In other words, 2^1 + ... + 2^N = 2^(N+1) - 2, for any natural number N.
Here is an example: let's say that we are interested in what you get when you add up the first N powers of 2. We experiment by calculating the first few:
2^1 = 1.
2^1 + 2^2 = 2 + 4 = 6.
2^1 + 2^2 + 2^3 = 2 + 4 + 8 = 14.
2^1 + 2^2 + 2^3 + 2^4 = 2 + 4 + 8 + 16 = 30.
Well, is there any obvious pattern here? In other words, is there a nice formula which will quickly tell us what (2^1 + ... + 2^N) is, for any natural number N?
Here is the magic trick: we simply define S(N) = 2^1 + ... + 2^N, for any natural number N, and we say the following magic words:
Quote:
S(N) = 2^1 + ... + 2^N.
So 2*S(N) = 2*(2^1 + ... + 2^N) = 2^2 + ... + 2^(N+1).
So S(N) = 2*S(N) - S(N) = (2^2 + ... + 2^(N+1)) - (2^1 + ... + 2^N) = 2^(N+1) - 2^1.
So 2*S(N) = 2*(2^1 + ... + 2^N) = 2^2 + ... + 2^(N+1).
So S(N) = 2*S(N) - S(N) = (2^2 + ... + 2^(N+1)) - (2^1 + ... + 2^N) = 2^(N+1) - 2^1.
In other words, 2^1 + ... + 2^N = 2^(N+1) - 2, for any natural number N.
There is a well-known generalization of this "trick", the sum of a geometric series.
r^0 + r^1 + r^2 + ... + r^n = ( r^(n+1) - 1 ) / ( r - 1 )
I discovered this formula when I was bored with the regular curriculum in high school math class. It was things like this that caused my otherwise rather strict teacher not to care whether or not I did my math homework
01 May 2012, 9:11 am
Species5618 wrote:
There is a well-known generalization of this "trick", the sum of a geometric series.
r^0 + r^1 + r^2 + ... + r^n = ( r^(n+1) - 1 ) / ( r - 1 )
I discovered this formula when I was bored with the regular curriculum in high school math class. It was things like this that caused my otherwise rather strict teacher not to care whether or not I did my math homework
r^0 + r^1 + r^2 + ... + r^n = ( r^(n+1) - 1 ) / ( r - 1 )
I discovered this formula when I was bored with the regular curriculum in high school math class. It was things like this that caused my otherwise rather strict teacher not to care whether or not I did my math homework
Nice! The formula for powers of 2 was obvious to me, having worked with binary a fair bit, but I didn't know the general formula. I'd probably replace the r^0 with 1 to make it more intuitive and easier to remember, ie.
r^1 + r^2 + ... + r^n = ( r^(n+1) - 1 ) / ( r - 1 ) - 1
