Summation /polynomials

here are the solutions:


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∑{k=2n,4n} (k^2)
=∑{k=1,4n} (k^2) - ∑{k=1,2n-1} (k^2)
=(4n)((4n)+1)((2(4n)+1)/6 - (2n-1)((2n-1)+1)(2(2n-1)+1)/6
=(4n)(4n+1)(8n+1)/6 - (2n-1)(2n)(4n-1)/6
=(112n³+60n ²+2n)/6

Now ∑{k=1,n} f(k)=(112n³+60n²+2n)/6 .......(1)
and (112n³+60n²+2n)/6 is of degree 3,
∴f(k) is of degree 2 and we may let f(k)=ak²+bk+c.
∑{k=1,n} f(k)
=∑{k=1,n}(ak²+bk+c)
=a∑{k=1,n}k² + b∑{k=1,n}k + c∑{k=1,n}1
=an(n+1)(2n+1)/6 + bn(n+1)/2 + cn
=a(2n³+3n²+n)/6 + b(n²+n)/2 + cn
=[(2an³ + (3a+3b)n² + (a+3b+6c)n]/6 .........(2)
Equating coefficients of (1) and (2):
2a=112
3a+3b=60
a+3b+6c=2
Solving, a=56, b=-36, c=9
∴f(k)=56k²-36k+9

∑{k=2n,4n} (k²) = ∑{k=1,n} f(n)
∑{k=2n,4n} (k²) = f(n) (∑{k=1,n} 1) as f(n) is independent of k
∑{k=2n,4n} (k²) = n* f(n)

f(n) = [∑{k=2n,4n} (k²)]/n