If Dr=⎛⎜⎝2r−12.3r−14.5r−1αβγ2n−13n−15n−1∣∣ ∣ ∣∣,then the value of ∑nr=1 Dr is
0
αβγ
α+β+γ
α−2n+β−3n+γ−4n
∑nr=1 2r−1=1+2+22+...+2n−1=2n−12−1=2n−1∑nr=1 2.3r−1=2(3n−1)3−1=3n−1∑nr=1 4.5r−1=4(5n−1)5−1=5n−1⇒∑nr=1 Dr=⎛⎜⎝2n−13n−15n−1αβγ2n−13n−15n−1∣∣ ∣ ∣∣=0