Akshay
Last Activity: 9 Years ago
∑k2 = [2n3 + 3n2+n]/6 call it eq.(1)
∑k = [n2+n]/2 call it eq.(2)
∑1 = n call it equation (3)
Note all limits are 1 to n.
6*eq(1) + 12*eq(2) + 6*eq(3) will give you,
∑6*(k2+2k+1)= 2n3+9n2+13n,
So, l(k) = 6*(k+1)2,
root[l(k)] = root(6) * (k+1).
just use equation 2 and 3 to solve for ∑root(l(k))