I'm sorry I don't have the book with me (don't know the problem number) since I moved, but I still remember the problem.
It said something about finding $\int x^n\ dx$ by letting $p_n = \int_0^1 x^n\ dx$ and finding a general formula for $p_n$.
$$\begin{align} \int_0^{2a} x^n\ dx &= 2a\int_0^1 (2ax)^n\ dx\\ &= (2a)^{n+1} p_n \end{align}$$
Additionally
$$\begin{align} (2a)^{n+1} p_n &= \int_0^{2a} x^n\ dx\\ &= \int_{-a}^a (x+a)^n\ dx\\ &= \int_{-a}^a \sum_{k=0}^n {n \choose k} x^k a^{n-k}\ dx\\ &= \sum_{k=0}^n {n \choose k} a^{n-k} \int_{-a}^a x^k\ dx\\ &= \sum_{k\text{ even}}^n {n \choose k} a^{n-k} 2\int_0^a x^k\ dx\\ &= \sum_{k\text{ even}}^n {n \choose k} 2a^{n-k} \cdot a^{k+1} p_k\\ &= 2a^{n+1} \sum_{k\text{ even}}^n {n \choose k} p_k\\ 2^n p_n &= \sum_{k\text{ even}}^n {n \choose k} p_k \end{align}$$
But this is where I'm stuck. I went through values of $n$ to confirm $p_n = p_0/(n+1)$ where $p_0 = 1$. But I haven't been able to prove this using induction (with step size of $1$ or $2$) or any other tricks I know. I thought I'd try rewriting the above to
$$\sum_{k\text{ even}}^n {n \choose k} (p_k-2p_n) = 0$$
But that doesn't help either. The problem always seems to boil down to not being able to simplify $\sum a\cdot b$.
$\endgroup$ 51 Answer
$\begingroup$Note that $2^n = (1+1)^n =\sum\limits_{k = 0}^n {n \choose k}$ and $0 = (1-1)^n =\sum\limits_{k = 0}^n {n \choose k}(-1)^k$, which combine to give you $\sum\limits_{k \text{ even}}^n {n \choose k} = 2^{n-1}$, and therefore $\sum\limits_{k \text{ odd}}^n {n \choose k} = 2^{n-1}$. This latter expression can be rewritten as $2^n = \sum\limits_{k \text{ even}}^{n+1} {n+1 \choose k+1} = \sum\limits_{k \text{ even}}^{n+1} {n \choose k}\frac{n+1}{k+1}$. Or: $$2^n\frac{1}{n+1} = \sum\limits_{k \text{ even}}^{n+1} {n \choose k}\frac{1}{k+1}$$
$\endgroup$ 3
