Question
If Dk=∣∣
∣
∣∣1nn2kn2+n+1n2+n2k−1n2n2+n+1∣∣
∣
∣∣ and ∑nk=1Dk=56, then n equals
∣
∣∣1nn2kn2+n+1n2+n2k−1n2n2+n+1∣∣
∣
∣∣ and ∑nk=1Dk=56, then n equals
Answer: Option D
:
D
∴∑nk=1Dk=∣∣
∣
∣∣∑nk=11nn2∑nk=1kn2+n+1n2+n2∑nk=1k−∑nk=11n2n2+n+1∣∣
∣
∣∣
=∣∣
∣
∣∣nnnn2+nn2+n+1n2+nn2n2n2+n+1∣∣
∣
∣∣=56
=Applying C2→C2→C1→C3→C3−C1, then
∣∣
∣∣n00n2+n10n0n+1∣∣
∣∣=56⇒n(n+1)=56=7×8⇒n=7
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:
D
∴∑nk=1Dk=∣∣
∣
∣∣∑nk=11nn2∑nk=1kn2+n+1n2+n2∑nk=1k−∑nk=11n2n2+n+1∣∣
∣
∣∣
=∣∣
∣
∣∣nnnn2+nn2+n+1n2+nn2n2n2+n+1∣∣
∣
∣∣=56
=Applying C2→C2→C1→C3→C3−C1, then
∣∣
∣∣n00n2+n10n0n+1∣∣
∣∣=56⇒n(n+1)=56=7×8⇒n=7
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