Find $\displaystyle \binom{n}{0}-\binom{n}{1}\frac{1}{4}+\binom{n}{2}\frac{1}{7}+\cdots \cdots $
What I tried:
the sum is $$\sum^{n}_{r=0}(-1)^r\binom{n}{r}\frac{1}{3r+1}$$
$$\sum^{n}_{r=0}(-1)^r\binom{n}{r}\int^{1}_{0}x^{3r}dx$$
$$\int^{1}_{0}\sum^{n}_{r=0}(-1)^r\binom{n}{r}x^{3r}dx$$
How do I solve? Help me, please!
On
From the Melzak's identity $$\sum_{k=0}^{j}\dbinom{j}{k}\left(-1\right)^{k}\frac{f\left(y-k\right)}{k+x}=\frac{f\left(x+y\right)}{x\dbinom{j+x}{j}}$$ where $f$ is an algebraic polynomial up to degree $j$ and $x\neq-k$ we get, taking $f\left(y\right)\equiv1$ and $x=1/a,\,a>0$, that $$\frac{1}{a}\sum_{k=0}^{j}\dbinom{j}{k}\frac{\left(-1\right)^{k}}{k+1/a}=\color{red}{\dbinom{j+1/a}{j}^{-1}}.$$ In your case, $a=3.$
As Zubin Mukerjee commented, the sum should be from $0$ to $n$ (as $\tbinom n {n+k}=0$ for $k>0$), and the fraction is an expression in terms of the iterator, $r$.
$$\sum_{r=0}^n \dfrac{(-1)^r\binom nr}{3r+1}\quad=\quad\int_0^1\sum_{r=0}^n(-1)^r\binom nr x^{3r}~\mathsf d x$$
Up next, the Binomial Expansion Theorem says: $$(1+a)^n = \sum_{r=0}^n a^r\binom nr$$
Take it from here.