JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Self Evaluation Test - Principle of Mathematical Induction

Using mathematical induction, the numbers \[{{a}_{n}}'s\]are defined by \[{{a}_{0}}=1,\,\,{{a}_{n+1}}=3{{n}^{2}}+n+{{a}_{n'}}\] \[(n\ge 0).\]Then, \[{{a}_{n}}\] is equal to

A) \[{{n}^{3}}+{{n}^{2}}+1\]

B) \[{{n}^{3}}-{{n}^{2}}+1\]

C) \[{{n}^{3}}-{{n}^{2}}\]

D) \[{{n}^{3}}+{{n}^{2}}\]

Correct Answer: B

Solution :

[b] Given, \[{{a}_{0}}=1,{{a}_{n+1}}=3{{n}^{2}}+n+{{a}_{n}}\]

\[\Rightarrow {{a}_{1}}=3(0)+0+{{a}_{0}}=1\]

\[\Rightarrow {{a}_{2}}=3{{(1)}^{2}}+1+{{a}_{1}}=3+1+1=5\]

For option (b),

Let \[P(n)={{n}^{3}}-{{n}^{2}}+1\]

\[\therefore \,\,\,\,\,P(0)=0-0+1=1={{a}_{0}}\]

\[P(1)={{1}^{3}}-{{1}^{2}}+1=1={{a}_{1}}\]

and \[P(2)={{(2)}^{3}}-{{(2)}^{2}}+1=5={{a}_{2}}\]

\[\therefore \,\,\,\,\,{{a}_{n}}={{n}^{3}}-{{n}^{2}}+1\]