Engineering
Mathematics
Question

If Dk=1nn2kn2+n+1n2+n2k1n2n2+n+1 and k=1nDk=56 then n equals

4
8
6
none of these
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Solution
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Dk=1nn2kn2+n+1n2+n2k1n2n2+n+1
k=1nDk=k=1n1nn2k=1nkn2+n+1n2+n2k=1nk1n2n2+n+1
56=nnnn(n+1)n2+n+1n2+nn2+n1n2n2+n+1
Applying C1 → C1 – C2 and C2 → C2 – C3
56=n00111n2+nn1(n+1)n2+n+1
⇒ 2n = 56
⇒ n = 28