If Dk=1nn2kn2+n+1n2+n2k−1n2n2+n+1 and ∑k=1n Dk=56, then n equals

∑k=1n Dk=56⇒∑k=1n 1nn∑k=1n 2kn2+n+1n2+n∑k=1n (2k−1)n2n2+n+1=56or nnnn(n+1)n2+n+1n2+nn2n2n2+n+1=56Applying C3→C3−C1 and C2→C2−C1, we getn00n(n+1)10n20n+1=56or n(n+1)=56 or n=7