(iii) 2+(2+5)+(2+5+8)+⋯+n2(3n+1)=n4(n+1)22 + (2+5) + (2+5+8) + \cdots + \dfrac{n}{2}(3n+1) = \dfrac{n}{4}(n+1)^22+(2+5)+(2+5+8)+⋯+2n(3n+1)=4n(n+1)2