As the two capacitors are connected in series, then
i = i1 +i2 where, i1 and i2 are the currents in the capacitors,
For a capacitor i= C dV/dt, where V is the voltage on the parallel combination
For i=0, (C1+ C2) dV/dt =0,
Integrating we get:(C1+ C2)V = constant= Total stored charges on the capacitors= qi1 +qi2 where qi1 and qi2 are the charges stored on capacitor 1 and 2 respectively. It turns out that the initial voltage= (qi1 +qi2)/(C1+C2)
This is expected results since the total capacitance of the two capacitors connected in parallel will be C1+C2 and the total initial charge on them will be q1+q2, so the initial voltage will be as given above.