If Dr=∣∣
∣
∣∣2r−12(3r−1)4(5r−1)xy3z2n−13n−15n−1∣∣
∣
∣∣ then prove that n∑r=1Dr=0.
Open in App
Solution
n∑r=12r−1=1+2+22+23+.............2n−1 = 1.(2n−1)2−1=2n−1, ∑2(3r−1)=2(1+3+32+..........+3n−1) =2.1.(3n−1)3−1=3n−1, ∑4(5r−1)=4(1+5+52+......+5n−1) =4.1(5n−1)5−1=5n−1. We have used the formula for G.P. Sn=a(rn−1)r−1 (r >1) Rest is as in Q.16