Find a closed-form expression for $$\binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots ,$$ where $n > 1$. You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful.
I really can't use the identity, as there is a floor function involved in my interpretation. How I see it, the pattern depends on the parity of $n$, but I'm probably wrong.
Can I get a solution?
HINT: There is no floor function involved. Your sum is
$$\sum_{k\ge 0}(2k+1)\binom{n}{2k+1}\;,$$
where it’s not necessary to specify an upper limit of summation, because $\binom{n}{2k+1}=0$ whenever $2k+1>n$. Now apply the identity and factor out the $n$:
$$\sum_{k\ge 0}(2k+1)\binom{n}{2k+1}=\sum_{k\ge 0}n\binom{n-1}{2k}=n\sum_{k\ge 0}\binom{n-1}{2k}\;.$$
The last summation is simply
$$\binom{n-1}0+\binom{n-1}2+\binom{n-1}4+\ldots\;;$$
$\binom{n-1}{2k}$ is the number of subsets of $\{1,\ldots,n-1\}$ with $2k$ elements, so that last summation is the number of subsets of $\{1,\ldots,n-1\}$ whose cardinalities are ... what kind of number? And how many of these are there?
Just extending Brian M. Scott's answer here.
From the binomial theorem:
$$(1 + x)^{n-1} = \sum_{i=0}^{n-1} \binom{n-1}{i}x^i$$ $$(1 - x)^{n-1} = \sum_{i=0}^{n-1} \binom{n-1}{i}(-x)^i$$
$$\begin{align} (1 + x)^{n-1} + (1 - x)^{n-1} & = \sum_{i=0}^{n-1} \binom{n-1}{i}x^i + \sum_{i=0}^{n-1} \binom{n-1}{i}(-x)^i\\ & = \sum_{i=0}^{n-1} \binom{n-1}{i}(x^i + (-x)^i)\\ & = \sum_{0 \le i \le n-1, i \text{ even}}^{n-1} \binom{n-1}{i}2x^i\\ \frac{(1 + x)^{n-1} + (1 - x)^{n-1}}{2} & = \sum_{0 \le i \le n-1, i \text{ even}}^{n-1} \binom{n-1}{i}x^i\\ \end{align}$$
Plug in $x = 1$ and the rhs becomes equivalent to Brian's last expression. So plug $x = 1$ into the left side and we get $\frac{2^{n-1}}{2} = 2^{n-2}$. The sum of the coefficients of the even terms in a binomial expression is equal to the sum of the coefficients of the odd terms.
