n∑r=1Dr=∣∣
∣
∣∣∑nr=12r−1∑nr=12(3r−1)∑nr=14(5r−1)xyz2n−13n−15n−1∣∣
∣
∣∣
=∣∣
∣
∣∣1+2+...2n−12{1+3....3n−1}4(1+5+....5n−1)xyz2n−13n−15n−1∣∣
∣
∣∣
=∣∣
∣
∣∣2n−12−12(3n−1)3−14(5n−1)5−1xyz2n−13n−15n−1∣∣
∣
∣∣=∣∣
∣
∣∣2n−13n−15n−1xyz2n−13n−15n−1∣∣
∣
∣∣=0
Since R1=R3