Using mathematical induction, the numbers $${a_n} 's$$ are defined $${a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \geqslant 0} \right).$$       Then, $$a_n$$ is equal to

Solution :
$$\eqalign{ & {\text{Given, }}{a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n} \cr & \Rightarrow {a_1} = 3\left( 0 \right) + 0 + {a_0} = 1 \cr & \Rightarrow {a_2} = 3{\left( 1 \right)^2} + 1 + {a_1} = 3 + 1 + 1 = 5 \cr & {\text{From option }}\left( B \right), \cr & {\text{Let, }}P\left( n \right) = {n^3} - {n^2} + 1 \cr & \therefore P\left( 0 \right) = 0 - 0 + 1 = 1 = {a_0} \cr & P\left( 1 \right) = {1^3} - {1^2} + 1 = 1 = {a_1} \cr & {\text{and }}P\left( 2 \right) = {\left( 2 \right)^3} - {\left( 2 \right)^2} + 1 = 5 = {a_2} \cr & \therefore {a_n} = {n^3} - {n^2} + 1 \cr} $$