Binomial coefficients

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August 19th 2011, 11:59 AM
alexmahone

Binomial coefficients

Prove that for all positive integers n > 1,

$\sum_{k=0}^n\frac{1}{k+1}\dbinom{n}{k}(-1)^{k+1}=0$

My attempt:

$(1+x)^n=\sum_{k=0}^n\dbinom{n}{k}x^k$

Integrating both sides,

$\frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\dbinom{n}{k} \frac{x^{k+1}}{k+1}+C$

Substituting x = 0,

$C=\frac{1}{n+1}$

$\frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\dbinom{n}{k} \frac{x^{k+1}}{k+1}+\frac{1}{n+1}$

Substituting x = -1,

$0=\sum_{k=0}^n\dbinom{n}{k} \frac{(-1)^{k+1}}{k+1}+\frac{1}{n+1}$

$\sum_{k=0}^n\frac{1}{k+1}\dbinom{n}{k}(-1)^{k+1}=-\frac{1}{n+1}$

What am I doing wrong?
August 19th 2011, 12:17 PM
alexmahone

Re: Binomial coefficients

On second thoughts, I'm beginning to think that the book may be wrong and I may be right.
August 19th 2011, 12:18 PM
Opalg

Re: Binomial coefficients

You are right and the question is wrong. You can check this by testing the formula with n=2. In that case, the sum becomes

$\sum_{k=0}^2\frac{1}{k+1}\dbinom{2}{k}(-1)^{k+1} = 1(-1) + \tfrac22(1) +\tfrac13(-1) = -\tfrac13.$

That agrees with your answer $-\tfrac{1}{n+1}$ (with n=2), but not with the given answer 0.

My guess is that the person who set this question forgot the constant when doing the integration. Congratulations to you for remembering that this is necessary. (Rock)
August 19th 2011, 12:34 PM
alexmahone

Re: Binomial coefficients

Quote:

Originally Posted by chisigma

The correct procedure should be...

$(1+t)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ t^{k}$ (1)

... then integrating both terms from 0 to x...

$\frac{(1+x)^{n+1}}{n+1} = \sum_{k=0}^{n} \binom{n}{k}\ \frac{x^{k+1}}{k+1}$ (2)

... then setting $x=-1$ ...

$\sum_{k=0}^{n} \binom{n}{k}\ \frac{(-1)^{k+1}}{k+1} =0$ (3)

Kind regards

$\chi$ $\sigma$

After you integrated from 0 to x, you did not substitute for the lower limit 0.
August 19th 2011, 12:40 PM
chisigma

Re: Binomial coefficients

Quote:

Originally Posted by alexmahone

After you integrated from 0 to x, you did not substitute for the lower limit 0.

I discovered that and I canceled my post before reading Your post...

Kind regards

$\chi$ $\sigma$