(v) 1×3+2×5+3×7+⋯+n×(2n+1)=n(n+1)(4n+5)61\times 3 + 2\times 5 + 3\times 7 + \cdots + n\times(2n+1) = \dfrac{n(n+1)(4n+5)}{6}1×3+2×5+3×7+⋯+n×(2n+1)=6n(n+1)(4n+5)