Let Dk=|12k2k-1nn2+n+2n2nn2+nn2+n+2|. If ∑k=1nDk=96, then n is equal to _________ . [2023]
(6)
Given, Dk=|12k2k-1nn2+n+2n2nn2+nn2+n+2|
∴ ∑k=1nDk=|∑k=1n1∑k=1n2k∑k=1n(2k-1)nn2+n+2n2nn2+nn2+n+2|
Apply ∑k=1n2k=2n(n+1)2=n(n+1)
∑k=1n(2k-1)=∑k=1n2k-∑k=1n1=2n(n+1)2-n=n2+n-n=n2
∴ ∑k=1nDk=|nn2+nn2nn2+n+2n2nn2+nn2+n+2|
Apply R2→R2-R1 and R3→R3-R1
⇒ |nn2+nn202000n+2|=96
⇒n(2(n+2)-0)=96⇒n(n+2)=48
⇒n2+2n-48=0⇒n2+8n-6n-48=0⇒n(n+8)-6(n+8)=0⇒(n-6)(n+8)=0
⇒n=6,-8
Since n cannot be negative, ∴ n=6