1.2+4+7+11+16+⋯2 + 4 + 7 + 11 + 16 + \cdots2+4+7+11+16+⋯ to nnn terms equalsa.16(n2+3n+8)\dfrac{1}{6}(n^2 + 3n + 8)61(n2+3n+8)b.n6(n2+3n+8)\dfrac{n}{6}(n^2 + 3n + 8)6n(n2+3n+8)c.16(n2−3n+8)\dfrac{1}{6}(n^2 - 3n + 8)61(n2−3n+8)d.n6(n2−3n+8)\dfrac{n}{6}(n^2 - 3n + 8)6n(n2−3n+8)Login to continueOnly logged in users canattempt or see the solution.
