sum of $\sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$

Question

Finding sum of $\displaystyle \sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$

Try: Using

$$\int^{1}_{0}x^m\cdot (1-x)^ndx = \frac{m!\cdot n!}{(m+n+1)!}=\frac{1}{(m+n+1)\binom{m+n}{n}}.$$

So $$ \frac{1}{(k+4)\binom{k+3}{k}}=\int^{1}x^3\cdot (1-x)^kdx$$

So our sum is $$\sum^{n}_{k=0}(-1)^k(k+4)\binom{n}{k}\int^{1}_{0}x^3(1-x)^kdx$$

$$ = \int^{1}_{0}x^3\sum^{n}_{k=0}(-1)^kk\binom{n}{k}(1-x)^kdx+4\int^{1}_{0}x^3\sum^{n}_{k=0}(-1)^k\binom{n}{k}(1-x)^k$$

$$ = -n\int^{1}_{0}x^{n+2}(1-x)dx+4\int^{1}_{0}x^{n+3}dx$$

$$ = -n\bigg[\frac{1}{n+3}-\frac{1}{n+4}\bigg]+\frac{4}{n+4} = -\frac{n}{(n+3)(n+4)}+\frac{4}{n+4} = $$

but answer given in book as $\displaystyle \frac{3}{n+3}$

Could some help me how to solve it, Thanks

Answer 1

I think $$\frac{4}{n+4}-\frac{n}{(n+3)(n+4)}=\frac{4(n+3)-n}{(n+3)(n+4)}=\frac{3(n+4)}{(n+3)(n+4)}=\frac{3}{n+3}$$

Answer 2

I think it is simpler to utilize binomial identities:

$$\dfrac{\binom nk}{\binom{k+3}k}=\dfrac{n!3!}{(n-k)!(k+3)!}=\dfrac6{(n+1)(n+2)(n+3)}\binom{n+3}{k+3}$$

$$\sum_{k=0}^n\dfrac{\binom nk}{\binom{k+3}k}=\dfrac6{(-1)^3(n+1)(n+2)(n+3)}\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$

$$\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$

$$=\sum_{r=0}^{n+3}\binom{n+3}r(-1)^r-\sum_{r=0}^2\binom{n+3}r(-1)^r$$

$$=(1-1)^n-\sum_{k=0}^2\binom{n+3}k(-1)^k=?$$