Let Δr=2r−123r−145r−1αβγ2n−13n−15n−1 , for r=1,2,…,n.r .Then ∑r=1n Δr

∑r=1n Δr=abcαβγ2n−13n−15n−1where a=∑r=1n 2r−1=2n−1b=∑r=1n 23r−1=23n−13−1=3n−1and c=∑r=1n 45r−1=45n−15−1=5n−1Thus, ∑r=1n Δr=abcαβγabc=0