If the equation (z+1)7+z7=0 has roots z1,z2,⋯z7 then what is the value of ∑r=17R(Zr)?

Step 1
I would choose a different approach, saving some labour:
The polynomial:
$p(z)=(z+1{)}^7+z^7=2z^7+7z^6+\cdots =2(z^7+\frac{7}{2z^6}+\cdots )$
And may also be written as:
$p(z)=2q(z)=2\prod _{k=1}^7(z-z_k)$
The sum of the roots of the monomial $q(z)$ is equal to minus the coefficient of $z^6$, as can be seen by expanding $q(z)$
The sum of the roots of $q(z)$ equals the sum of roots of $p(z)$:
$Re(\sum _{k=1}^7{z}_k)=(\sum _{k=1}^7{z}_k)=-\frac{7}{2}$

Step 1
You can observe that z=0 is not a solution, so the equation can be rewritten as
$(1+\frac{1}{z}{)}^7=-1$
Hence we get
$1+\frac{1}{z_k}=w_k$
for $1\le k\le 7$, where $w_k=\exp ((\pi +2k\pi )\frac{i}{7})$
Write ${\alpha }_k=\frac{\pi +2k\pi }{14}$, so we have
$z_k=\frac{1}{e^{2{\beta }_{k}i}-1}=\frac{e^{-{\beta }_{k}i}}{2i\sin {\beta }_k}$
and therefore
$z_k=-\frac{i}{2\sin {\beta }_k}(\cos {\beta }_k-i\sin {\beta }_k)$
and we obtain $Re{z}_k=\frac{1}{2}$