Q91 / 180 · COMEDK 2023 (Memory Based) · 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

Given, $a_0=1, a_{n+1}=3 n^2+n+a_n$ $\begin{array}{ll} \Rightarrow & a_1=3(0)^2+(0)+a_0=0+0+1=1 \\ \Rightarrow & a_2=3(1)^2+(1) a_1=3+1+1=5 \end{array}$ From option (b), Let $P(n)=n^3-n^2+1$ $\begin{aligned} & P(0)=(0)^3-(0)^2+1=1=a_0 \\ & P(1)=(1)^3-(1)^2+1=1-1+1=a_1 \\ & P(2)=(2)^3-(2)^2+1=8-4+1=5=a_2 \end{aligned}$ Thus, $a_n=n^3-n^2+1$

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

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