Combinatronics Summation Series I have a series as follows:

$$\sum_{k=1}^{15} {30 \choose 2k-1}/2k$$

I have tried expanding and evaluating by replacing $${30\choose29}/30 $$ with $${30\choose1}/30 $$
How do you solve it, is there a general approach to attempt such questions?
 A: 
We  obtain
  \begin{align*}
\color{blue}{\sum_{k=1}^{15}\binom{30}{2k-1}\frac{1}{2k}}&=\frac{1}{31}\sum_{k=1}^{15}\binom{31}{2k}\tag{1}\\
&=\frac{1}{31}\sum_{k=1}^{15}\binom{31}{32-2k}\tag{2}\\
&=\frac{1}{31}\sum_{k=1}^{15}\binom{31}{2k-1}\tag{3}\\
&\color{blue}{=\frac{1}{31}\left(2^{30}-1\right)}\tag{4}
\end{align*}

Comment:


*

*In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

*In (2) we  change the  order of summation $k\rightarrow15-k+1=16-k$.

*In (3) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (4) we observe  that the series in (1)  and  (3) summed up give
$\sum_{k=1}^{30}\binom{31}{k}=2^{31}-2$.
A: $$\frac1{2k}=\int_0^1x^{2k-1}\,dx.$$
Your sum is
$$\int_0^1\left(\sum_{k=1}^{15}\binom{30}{2k-1}x^{2k-1}\right)\,dx.$$
Now
$$\sum_{k=1}^{15}\binom{30}{2k-1}x^{2k-1}$$
is the sum of the terms with odd powers of $x$ in
$$\sum_{j=0}^{30}\binom{30}{j}x^{j}$$
and there's a standard trick to extracting the "odd part" of a polynomial
or power series...
