# Binomial coefficients 3

• Sep 16th 2009, 07:49 PM
manusform
Binomial coefficients 3
This problem is very difficult...can someone help me with this please? ...this is all about binomials

1. Show that

$
\sum_{k=1}^{n}k\binom{n}{2k+1}=(n-2){2}^{n-3}
$
• Sep 18th 2009, 02:31 PM
Taluivren
Hello,
$(1-x)^n=\sum_{k=0}^n (-1)^k \binom{n}{k}x^k=-\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1} + \sum_{k=0}^n \binom{n}{2k}x^{2k}$

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

Take derivative with respect to $x$ of both equalities

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

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

Set $x=1$ in all four equalities and you get

$\sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} =2^{n-1}$

$\sum_{k=0}^n (2k+1) \binom{n}{2k+1} = \sum_{k=0}^n 2k \binom{n}{2k} = n2^{n-2}$ for $n>1$

Now because

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

you get

$\sum_{k=0}^n k \binom{n}{2k+1} = \frac{1}{2}(n2^{n-2} - 2^{n-1}) =\frac{1}{2}(n2^{n-2} - 2\cdot2^{n-2})=(n-2)2^{n-3}$ for $n>1$
• Sep 30th 2009, 11:42 AM
streethot
I think you made a mistake.

$\sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} \neq2^{n-1}$

The right form is:

$\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2k +1} = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2k} =2^{n-1}$
• Sep 30th 2009, 11:52 AM
streethot
Sorry, both two form are right.
• Sep 30th 2009, 12:02 PM
Taluivren
Hi streethot,

I don't think it is a mistake, since the definition of binomial coefficients says
$\binom{n}{k}=0$ for $k>n$. This means your form and my form are equivalent.

I gave a derivation of $\sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} = 2^{n-1}$ above.