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Question

# If Dk=∣∣ ∣ ∣∣1nn2kn2+n+1n2+n2k−1n2n2+n+1∣∣ ∣ ∣∣ and ∑nk=1Dk=56, then n equals ?

A
4
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B
6
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C
8
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D
none of these
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Solution

## The correct option is D none of theseDk=∣∣ ∣ ∣∣1nn2kn2+n+1n2+n2k−1n2n2+n+1∣∣ ∣ ∣∣∑nk=1Dk=∣∣ ∣ ∣∣∑nk=11nn∑nk=12kn2+n+1n2+n∑nk=1(2k−1)n2n2+n+1∣∣ ∣ ∣∣56=∣∣ ∣ ∣ ∣ ∣∣n+1nn2n(n+1)2n2+n+1n2+n2n(n+1)2−(n+1)n2n2+n+1∣∣ ∣ ∣ ∣ ∣∣56=∣∣ ∣ ∣∣n+1nnn2+nn2+n+1n2+nn2−1n2n2+n+1∣∣ ∣ ∣∣Applying C2→C2−C1,C3→C3−C156=∣∣ ∣ ∣∣1−1−1n2+n10n2−11n+2∣∣ ∣ ∣∣Applying R2→R2+R1,R3→R3+R156=∣∣ ∣ ∣∣1−1−1n2+n+10−1n20n+1∣∣ ∣ ∣∣56=−1((n+1)(n2+n+1)+n2)56=−1(n3+n2+n+n2+n+1+n2)n3+3n2+2n=−57

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