Evaluate the sum $\sum_{k=0}^n \frac{\binom n k}{(k+1)(k+3)} $

Question

How to sum the sequence $$\sum_{k=0}^n \frac{\binom n k}{(k+1)(k+3)} $$ where $n \choose k$ are the usual binomial coefficients in expansion of $(1+x)^n$.

I know how to sum such sequence when the denominator has consecutive multiplicands (by repeatedly integrating the expansion of $(1+x)^n$).

Does there even exist a general procedure for such sums?

Answer 1

One can proceed similarly as in Is there a closed formula for $\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}\binom{n}{k}$? : Integrating $$ \sum_{k=0}^n \binom n k x^k = (1+x)^n $$ gives $$ \sum_{k=0}^n \binom n k \frac{x^{k+1}}{k+1} = \frac{1}{n+1}\bigl((1+x)^{n+1} - 1 \bigr). $$ Now multiply by $x$ and then integrate again, for $0 \le x \le 1$: $$ \sum_{k=0}^n \binom n k \frac{1}{(k+1)(k+3)} = \frac{1}{n+1} \int_0^1 x\bigl((1+x)^{n+1} - 1 \bigr) \, dx \, . $$ The integral can be evaluated with integration by parts, the result is $$ \sum_{k=0}^n \binom n k \frac{1}{(k+1)(k+3)} = \frac{2^{n+3} - (n+4)}{2(n+2)(n+3)} \, . $$

Answer 2

Hint:

$$\dfrac{\binom nr}{(r+1)(r+3)}=\cdots=\dfrac1{(n+1)(n+2)(n+3)}\cdot\dfrac{(n+3)!(r+2)}{(r+3)!(n+3-(r+3))!}=?$$

Again,$\displaystyle\dfrac{(n+3)!(r+2)}{(r+3)!(n+3-(r+3))!}=\dfrac{(n+3)!(r+3-1)}{(r+3)!(n+3-(r+3))!}=\cdots=(n+3)\binom{n+2}{r+2}-\binom{n+3}{r+3}$

Answer 3

Hint:

Let $\displaystyle\dfrac{\binom nr}{(r+1)(r+3)}=a\binom{n+3}{r+3}+b\binom{n+2}{r+2}+c\binom{n+1}{r+1}$ where $a,b,c$ are arbitrary constants

On simplification, $$\dfrac1{(r+1)(r+3)}=\dfrac{a(n+3)(n+2)(n+1)}{(r+3)(r+2)(r+1)}+\dfrac{b(n+2)(n+1)}{(r+2)(r+1)}+\dfrac{c(n+1)}{r+1}$$

$$\iff r+2=a(n+3)(n+2)(n+1)+b(n+2)(n+1)(r+3)+c(n+1)(r+3)(r+2)$$

Set $r+3=0\implies -3+2=a(n+3)(n+2)(n+1)\implies a=?$

$r+2=0\implies 0=-1+b(n+2)(n+1)(-2+3)\iff b=?$

Comparing the coefficients of $r^2, c=0$