Decreasing

[orosmatthew]

Sorry if I put this under the wrong category because I don't know where it goes

So I know that if you take a number like 100 and half it you get 50 then again 25 again 12.5, 6.25 etc.

If you half it enough times you can add them und you would get roughly 100 or whatever the number you started with.
I was wondering if instead of multiplying by 0.5 each time you could multiply it by 0.7 or something. When I multiply by 0.7 enough times and ad them up it doesn't roughly equal the number I started with.

ex.

by 0.5-----------by 0.7
100--------------100

50---------------70
25---------------49
12.5-------------34.3
6.25-------------24.01
3.125------------16.807

Total:96.875----Total:194.117

Why won't multiplying by 0.7 work?

 

[Deleted member 4993]

Sorry if I put this under the wrong category because I don't know where it goes

So I know that if you take a number like 100 and half it you get 50 then again 25 again 12.5, 6.25 etc.

If you half it enough times you can add them und you would get roughly 100 or whatever the number you started with.
I was wondering if instead of multiplying by 0.5 each time you could multiply it by 0.7 or something. When I multiply by 0.7 enough times and ad them up it doesn't roughly equal the number I started with.

ex.

by 0.5-----------by 0.7
100--------------100

50---------------70
25---------------49
12.5-------------34.3
6.25-------------24.01
3.125------------16.807

Total:96.875----Total:194.117

Why won't multiplying by 0.7 work?

Because you are creating a geometric series and you need to look at the formula for sum of the geometric series.

 

[orosmatthew]

Because you are creating a geometric series and you need to look at the formula for sum of the geometric series.

Ok, so how do I find the formula or what to search on Google or something?

 

[Ishuda]

Ok, so how do I find the formula or what to search on Google or something?

Consider
(x - 1) (xⁿ + x^n-1 + x^n-2 + .... + x + 1)
= (xⁿ⁺¹ + xⁿ + x^n-1 + x^n-2 + .... + x² + x)
- (xⁿ + x^n-1 + x^n-2 + .... + x² + x + 1)
= (xⁿ⁺¹ + xⁿ + x^n-1 + x^n-2 + .... + x² + x)
+\(\displaystyle \space \space\space \space \space \space\space\space\space\space\space\)(- xⁿ - x^n-1 - x^n-2 - ....\(\displaystyle \space \space\space\) - x² - x - 1)
= xⁿ⁺¹ - 1
or
\(\displaystyle x^n + x^{n-1} + ... + x + 1 = \frac{x^{n+1} - 1}{x - 1}\)

 

[orosmatthew]

Consider
(x - 1) (xⁿ + x^n-1 + x^n-2 + .... + x + 1)
= (xⁿ⁺¹ + xⁿ + x^n-1 + x^n-2 + .... + x² + x)
- (xⁿ + x^n-1 + x^n-2 + .... + x² + x + 1)
= (xⁿ⁺¹ + xⁿ + x^n-1 + x^n-2 + .... + x² + x)
+\(\displaystyle \space \space\space \space \space \space\space\space\space\space\space\)(- xⁿ - x^n-1 - x^n-2 - ....\(\displaystyle \space \space\space\) - x² - x - 1)
= xⁿ⁺¹ - 1
or
\(\displaystyle x^n + x^{n-1} + ... + x + 1 = \frac{x^{n+1} - 1}{x - 1}\)

Whats with the ... or is that just fake? :?:

 

[HallsofIvy]

Whats with the ... or is that just fake? :?:

If you are asking what the "..." means that is a standard notation of a sum "continuing in the same way"
For n= 3 this would be \(\displaystyle x^3+ x^2+ x+ 1\).
For n= 5 it would be \(\displaystyle x^5+ x^4+ x^3+ x^2+ x+ 1\)

For general n, it would be \(\displaystyle x^n+ x^{n-1}+ x^{n-1}+ \) and now how many terms we have will depend on n so we write \(\displaystyle \cdot\cdot\cdot+ x^3+ x^2+ x+ 1\) with the "\(\displaystyle \cdot\cdot\cdot\)" representing all of the terms in between.