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Question

If nk=1 km=1 m2 = an4 + bn3 + cn2 + dn + e. then


A

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B

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C

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D

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E

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Solution

The correct options are
A


B


C


D


nk=1 km=1 = an4 + bn3 + cn2 + dn + e -----------(1)

nk=1 km=1 m2 = nk=1 k(k+1)(2k+1)6

= 16 nk=1 [2k3+3k2+k]

Here, nk=1[k3] = n2(n+1)24

nk=1[k2] = n(n+1)(2n+1)6

nk=1[k] = n(n+1)2

=16 [2.n2(n+1)24+3.n(n+1)(2n+1)6+n(n+1)2]

= 16 12 n(n+1) [n(n+1) + 2n + 1 + 1]

= 112 n(n+1) [n2 + n + 2n + 1 + 1]

= 112 n [n3 + 3n2 + 2n + n2 + 3n + 2]

= n12 [n3 + 4n2 + 5n + 2]

= n412 + n33 + 512n2 + n6

Compare it with equation an4 + bn3 + cn2 + dn + e.

We get,

a = 112, b =13, c = 512, d = 16.


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