2. Solved Papers
  3. Manipal Engineering

Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    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 :

    Given, \[{{a}_{0}}=1,\,\,{{a}_{n+1}}=3{{n}^{2}}+n+{{a}_{n}}\] \[\Rightarrow \]               \[{{a}_{1}}=3(0)+0+{{0}_{0}}=1\] and        \[{{a}_{2}}=3{{(1)}^{2}}+1+{{a}_{1}}=3+1+1=5\] From 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        \[P(2)={{(2)}^{3}}-{{(2)}^{2}}+1=5={{a}_{2}}\] Hence, option (b) is correct.


You need to login to perform this action.
You will be redirected in 3 sec spinner