This problem is very difficult...can someone help me with this please? ...this is all about binomials
1. Show that
$\displaystyle
\sum_{k=1}^{n}k\binom{n}{2k+1}=(n-2){2}^{n-3}
$
This problem is very difficult...can someone help me with this please? ...this is all about binomials
1. Show that
$\displaystyle
\sum_{k=1}^{n}k\binom{n}{2k+1}=(n-2){2}^{n-3}
$
Hello,
$\displaystyle (1-x)^n=\sum_{k=0}^n (-1)^k \binom{n}{k}x^k=-\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1} + \sum_{k=0}^n \binom{n}{2k}x^{2k}$
$\displaystyle (1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k=\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1 } + \sum_{k=0}^n \binom{n}{2k}x^{2k}$
Take derivative with respect to $\displaystyle x$ of both equalities
$\displaystyle -n(1-x)^{n-1}=-\sum_{k=0}^n (2k+1) \binom{n}{2k+1}x^{2k} + \sum_{k=0}^n 2k \binom{n}{2k}x^{2k-1}$
$\displaystyle n(1+x)^{n-1}=\sum_{k=0}^n (2k+1) \binom{n}{2k+1}x^{2k} + \sum_{k=0}^n 2k \binom{n}{2k}x^{2k-1}$
Set $\displaystyle x=1$ in all four equalities and you get
$\displaystyle \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} =2^{n-1}$
$\displaystyle \sum_{k=0}^n (2k+1) \binom{n}{2k+1} = \sum_{k=0}^n 2k \binom{n}{2k} = n2^{n-2}$ for $\displaystyle n>1$
Now because
$\displaystyle \sum_{k=0}^n (2k+1) \binom{n}{2k+1} = 2\sum_{k=0}^n k \binom{n}{2k+1}+\sum_{k=0}^n \binom{n}{2k+1}$
you get
$\displaystyle \sum_{k=0}^n k \binom{n}{2k+1} = \frac{1}{2}(n2^{n-2} - 2^{n-1}) =\frac{1}{2}(n2^{n-2} - 2\cdot2^{n-2})=(n-2)2^{n-3}$ for $\displaystyle n>1$
I think you made a mistake.
$\displaystyle \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} \neq2^{n-1}$
The right form is:
$\displaystyle \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2k +1} = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2k} =2^{n-1}$
Hi streethot,
I don't think it is a mistake, since the definition of binomial coefficients says
$\displaystyle \binom{n}{k}=0 $ for $\displaystyle k>n$. This means your form and my form are equivalent.
I gave a derivation of $\displaystyle \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} = 2^{n-1}$ above.