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Question

If Dk=∣ ∣ ∣1nn2kn2+n+1n2+n2k1n2n2+n+1∣ ∣ ∣ and nk=1Dk=56, where nN, then n is equal to 


Solution

It is given that nk=1Dk=56
where, Dk=∣ ∣ ∣1nn2kn2+n+1n2+n2k1n2n2+n+1∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣nk=11nnnk=12kn2+n+1n2+nnk=1(2k1)n2n2+n+1∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣=56

nk=11=n
nk=12k=2(1+2++n)=2n(n+1)2=n(n+1)
nk=1(2k1)=nk=12knk=11=n(n+1)n=n2

∣ ∣ ∣nnnn(n+1)n2+n+1n2+nn2n2n2+n+1∣ ∣ ∣=56

C2C2C1, C3C3C1
∣ ∣ ∣n00n(n+1)10n20n+1∣ ∣ ∣=56

n(n+1)=56
n2+n56=0
n2+8n7n56=0
(n+8)(n7)=0
n=7,n=8 (Not Possible)
n=7


 

Mathematics

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