prove $\sum_{k=0}^{3n+2}\frac{(-1)^k}{6n+5-k}\binom{6n+5-k}{k}=\frac{1}{6n+5}$ I want to
prove that $$\sum_{k=0}^{3n+2}\frac{(-1)^k}{6n+5-k}\binom{6n+5-k}{k}=\frac{1}{6n+5}$$
Let $\displaystyle f_n(x)=\sum_{k=0}^{\lfloor (n+1)/2\rfloor}\binom{n-k+1}{k}x^{n-k}$,
 then we have a recurrence relation 
$$f_n(x)=xf_{n-1}(x)+xf_{n-2}(x) \quad ; \quad f_1(x)=x+1,\quad f_0(x)=1$$
Hence, $$\sum_{k=0}^{3n+2}\frac{(-1)^k}{6n+5-k}\binom{6n+5-k}{k}=\int_{-1}^0f_{6n+4}(x)dx$$
What will I do from here? Thank you.

Chebyshev polynomials of the second kind have the following closed form:
$$ U_n(x) = \sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}(-1)^r (2x)^{n-2r} \tag{1} $$
hence:
$$ x^{n-1}\,U_n(x/2) = \sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}(-1)^r (x)^{2n-2r-1} \tag{2} $$ 
and:
$$ S_n=\sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}\frac{(-1)^r}{n-r} = 2\int_{0}^{1}x^{n-1} U_n(x/2)\,dx. \tag{3} $$
By substituting $x=2\cos\theta$, the last integral turns into an elementary integral and we get:
$$\boxed{ \sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}\frac{(-1)^r}{n-r} = \color{red}{\frac{1}{n}}}\tag{4}$$
for any $n\equiv \pm1\pmod{6}$. As an alternative, we may use the generating function for Chebyshev polynomials of the second kind, by replacing $x$ with $x/2$, then $t$ with $xt$. $S_n$ then becomes the coefficient of $t^n$ in a simple analytic function.
