What will the following formula go ?

[Pattielli]

Can you point out how to simplify the following formula ?

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

Thank you

[matt grime]

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?

[Pattielli]

Yes, they are binomial coeffs.
I tried till n reaches 4 and I figured it out...
n=0 \frac{C^0_0}{x}=\frac{1}{x}
n=1 \frac{C^0_1}{x}-\frac{C^1_1}{x+1}=\frac{1}{x(x+1)}
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)}
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)}
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)}

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

Thank Matt so very much for your suggestions, :smile:

[matt grime]

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

[Pattielli]

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 learnt new things from you, Matt. :sm:

Thank Matt very much...