Solution
= Solution
Subtract successive terms and use a <telescoping series>:
$$
z_{n+1}-z_1=\sum_{j=1}^n(z_{j+1}-z_j)=\sum_{j=1}^nc_j.
$$
Thus the <linear recurrence relation> has solution
$$
\boxed{z_{n+1}=z_1+\sum_{j=1}^nc_j.}
$$
= Solution
Subtract successive terms and use a <telescoping series>:
$$
z_{n+1}-z_1=\sum_{j=1}^n(z_{j+1}-z_j)=\sum_{j=1}^nc_j.
$$
Thus the <linear recurrence relation> has solution
$$
\boxed{z_{n+1}=z_1+\sum_{j=1}^nc_j.}
$$