What will the following formula go ?

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In summary, the formula given is a simplified version of the binomial coefficients, where the coefficients are arranged in an alternating pattern depending on the value of n. This pattern can be observed by working out the coefficients for different values of n. The simplified formula can also be written as \frac{n!}{x(x+1)...(x+n)}.
  • #1
Pattielli
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Can you point out how to simplify the following formula ?

[tex]\frac{C^0_n}{x}-\frac{C^1_n}{x+1}+...+(-1)^n\frac{C^n_n}{x+n}[/tex]

Thank you
 
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  • #2
your Cs are binomial coeffs?

have you worked itout for the cases n=0,1,2? what did you get there? is there a pattern you can see?
 
  • #3
Thanks Matt for your help

Yes, they are binomial coeffs.
I tried till n reaches 4 and I figured it out...
[tex]n=0 \frac{C^0_0}{x}=\frac{1}{x}[/tex]
[tex]n=1 \frac{C^0_1}{x}-\frac{C^1_1}{x+1}=\frac{1}{x(x+1)}[/tex]
[tex]n=2, \frac{C^0_2}{x}-\frac{C^1_2}{x+1}+\frac{C^2_2}{x+2}=\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}=\frac{2}{x(x+1)(x+2)}[/tex]
[tex]n=3, \frac{C^0_3}{x}-\frac{C^1_3}{x+1}+\frac{C^2_3}{x+2}-\frac{C^3_3}{x+3}=\frac{6}{x(x+1)(x+2)(x+3)}[/tex]
[tex]n=4, \frac{C^0_4}{x}-\frac{C^1_4}{x+1}+\frac{C^2_4}{x+2}-\frac{C^3_4}{x+4}+\frac{C^4_4}{x+4}=\frac{24}{x(x+1)(x+2)(x+3)(x+4)}[/tex]

So I think it will be [tex]\frac{n!}{x(x+1)...(x+n)}[/tex]

Thank Matt so very much for your suggestions, :smile:
 
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  • #4
that seems about right:

let P=x(x+1)(x+2)...(x+n) and let P(r) be P, but where you omit the factor (x+r)

then you want to work out
{P(0) -P(1)nC1 + P(2)nC2 ...)/P

if you work out the coeff of x^s in the bracket you see lots of things happening:

x^(n-1) has coeff the alternating sum of all the binom coeffs, so it's zero,
x^0 is just the constant term in P(0), cos all the other terms P(s) have a factor of x in them. you should tidy up that to work for all coeffs
 
  • #5
matt grime said:
that seems about right:

let P=x(x+1)(x+2)...(x+n) and let P(r) be P, but where you omit the factor (x+r)

then you want to work out
{P(0) -P(1)nC1 + P(2)nC2 ...)/P

if you work out the coeff of x^s in the bracket you see lots of things happening:

x^(n-1) has coeff the alternating sum of all the binom coeffs, so it's zero,
x^0 is just the constant term in P(0), cos all the other terms P(s) have a factor of x in them. you should tidy up that to work for all coeffs
Oh Well, That is really great, I have just learned new things from you, Matt. :sm:

Thank Matt very much...
 

What will the following formula go ?

The answer to this question will depend on the specific formula in question. However, in general, formulas are used to represent mathematical relationships and perform calculations. They can be used in various fields such as physics, chemistry, and economics.

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The purpose of using a formula is to express a mathematical relationship in a concise and efficient manner. Formulas allow us to perform calculations and analyze data quickly and accurately.

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Interpreting a formula involves understanding the variables and symbols used, as well as the mathematical operations involved. It is important to carefully read and analyze the formula to determine its meaning and how it can be applied.

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Yes, formulas can be modified or adapted to fit specific situations or problems. However, it is important to understand the underlying mathematical principles and not make arbitrary changes that could alter the meaning or accuracy of the formula.

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Common mistakes when using formulas include incorrect input of values, using the wrong formula for a specific problem, and not considering all relevant variables. It is important to double check the inputs and ensure that the formula used is appropriate for the given problem.

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