We have, k=1∑16Dk=∣∣abc(k=1∑162k)3(k=1∑1644)7(k=1∑168k)216−12(416−1)4(816−1)∣∣ ⇒k=1∑16Dk=∣∣abc(2−1216−1)3×4(4−1416−1)7×8(8−1816−1)216−12(416−1)4(816−1)∣∣ ⇒k=1∑16Dk=∣∣abc2(216−1)4(416−1)8(816−1)216−12(416−1)4(816−1)∣∣ ⇒k=1∑16Dk=2∣∣abc(216−1)2(416−1)4(816−1)216−12(416−1)4(816−1)∣∣
[taking 2 common from C2 ] ⇒k=1∑16Dk=2×0=0