Combinatronics Summation Series

Question

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?

Answer 1

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 $\sum_{k=1}^{30}\binom{31}{k}=2^{31}-2$.

Answer 2

$$\frac1{2k}=\int_0^1x^{2k-1}\,dx.$$ 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...