I Thought These Were Partial Fractions

I am giving a solution by replacing $71$ by $n$

First define a rational function $R(t)=\sum_{j=1}^{n}\frac{x_{j}}{t+j}$

$R(t)=\frac{P(t)}{Q(t)}$ for some polynomials $P(t)$ and $Q(t)$ where $degP<n$

$Q(t)=(t+1)(t+2)....(t+n)$

Define $G(t)=P(t)(2t+1)-4Q(t)$

$G(t)=0$ for $t=1,.....,n$ and hence $G(t)=c(t-1)....(t-n)$ where $c$ is a constant.

By putting $t=-\frac{1}{2}$ show that $c=4(-1)^{n-1}\frac{1}{2n+1}$

Now its easy to show $\sum_{j=1}^{n}\frac{x_{j}}{2j+1}=\frac{1}{2}R(\frac{1}{2})=1-\frac{1}{(2n+1)^{2}}$

Now, $(2n+1)^{2}$ and $(2n+1)^{2}-1$ are coprime integers.

Hence,by putting $n=71$ ,we get $p+q=143^{2} +143^{2}-1=40897$