Q43 / 160 · TS EAMCET 2009 · Mathematics

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

  • A
    $n^3+n^2+1$
  • Correct answer
    B
    $n^3-n^2+1$
  • C
    $n^3-n^2$
  • D
    $n^3+n^2$

Solution

$\begin{aligned} & \text { Given, } a_0=1, a_{n+1}=3 n^2+n+a_n \\ & \Rightarrow \quad a_1=3(0)+0+a_0=1 \\ & \text { and } \quad a_2=3(1)^2+1+a_1=3+1+1=5 \end{aligned}$ From option (b), $\begin{array}{ll} \text { Let } & P(n)=n^3-n^2+1 \\ \therefore & P(0)=0-0+1=1=a_0 \\ \text { and } & P(1)=1^3-1^2+1=1=a_1 \\ & P(2)=(2)^3-(2)^2+1=5=a_2 \end{array}$ Hence, option (b) is correct.

Hence, the correct answer is option: 'B'.

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