If A = $\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$, then by the principle of Mathematical induction, prove that Aⁿ = $\left[\begin{array}{cc}1& 2\mathrm{n}\\ 0& 1\end{array}\right]$

The value of 2, 6, 10 ... (4n - 6)(4n - 2) is equal to

• C(2n, n)

• (n + 1)(n + 2)(n + 3) ... (2n)

• n! P (2n, n)

• None of above

$$\begin{gathered} \mathrm{Using} \mathrm{mathematical} \mathrm{induction}, \mathrm{the} \mathrm{numbers} {\mathrm{a}}_{\mathrm{n} }\text{'}\mathrm{s} \mathrm{are} \mathrm{defined} \mathrm{by}, \\ {\mathrm{a}}_0 = 1, {\mathrm{a}}_{\mathrm{n} + 1} = 3{\mathrm{n}}^2 + \mathrm{n} + {\mathrm{a}}_{\mathrm{n}}, \left(\mathrm{n} \ge 0\right) \\ \mathrm{Then} {\mathrm{a}}_{\mathrm{n}} = ? \end{gathered}$$

• n³ + n² + 1

• n³ - n² + 1

• n³ - n²

• n³ + n²