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