Exercise 1.4.26* (Conundrum 8/8)

Question

Show that
$$\sum_{k=1}^n k \binom{n}{2k+1} = (n-2)2^{n-3}.$$

Richard Ganaye · Accepted Answer

\begin{proof} Here we assume $n \geq 2$ (the formula is false for $n=1$).

Note that $$S = \sum_{k=1}^n k \binom{n}{2k+1} = \sum_{k=1}^{\lfloor (n-1)/2 floor} k \binom{n}{2k+1},$$
since $\binom{n}{2k+1} = 0$ if $ k > \lfloor (n-1)/2 floor$.

If we separate odd and even powers in the binomial formula, we obtain
\begin{align}
(1+x)^n &= \phantom{-}\sum_{k=0}^{\lfloor (n-1)/2 floor} \binom{n}{2k+1} x^{2k+1} +  \sum_{k=0}^{\lfloor n/2 floor} \binom{n}{2k} x^{2k},\
(1-x)^n &= -\sum_{k=0}^{\lfloor (n-1)/2 floor} \binom{n}{2k+1} x^{2k+1} +  \sum_{k=0}^{\lfloor n/2 floor} \binom{n}{2k} x^{2k}.
\end{align}
The difference of (1) and (2) gives
$$(1+x)^n - (1-x)^n = 2 \sum_{k=0}^{\lfloor (n-1)/2 floor} \binom{n}{2k+1} x^{2k+1}.$$
In particular, for $x=1$, $$T = \sum_{k=0}^{\lfloor (n-1)/2 floor} \binom{n}{2k+1} = 2^{n-1}.$$
The derivation of (1) and (2) gives
\begin{align}
\phantom{-} n(1+x)^{n-1} &= \phantom{-}\sum_{k=0}^{\lfloor (n-1)/2 floor} (2k+1)\binom{n}{2k+1} x^{2k} +  \sum_{k=1}^{\lfloor n/2 floor} 2k \binom{n}{2k} x^{2k-1},\
-n(1-x)^{n-1}&= -\sum_{k=0}^{\lfloor (n-1)/2 floor} (2k+1)\binom{n}{2k+1} x^{2k} +  \sum_{k=1}^{\lfloor n/2 floor} 2k\binom{n}{2k} x^{2k-1}.
\end{align}
The difference of (3) and (4) gives
\begin{align*}
n[(1+x)^{n-1} + (1-x)^{n-1}] &= 2 \sum_{k=0}^{\lfloor (n-1)/2 floor} (2k+1)\binom{n}{2k+1} x^{2k}.
\end{align*}
For $x = 1$, using $n\geq 2$, we obtain
\begin{align*}
n2^{n-1} &= 2 (2S + T)\
&=2^2S + 2^{n}.
\end{align*}
Therefore
$$ S = \sum_{k=1}^n k \binom{n}{2k+1} = (n-2)2^{n-3}.$$
\end{proof}

Exercise 1.4.26* (Conundrum 8/8)

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Exercise 1.4.26* (Conundrum 8/8)

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