Q63 / 200 · Manipal (MET) 2010 · Mathematics

The numbers $a_n^{\prime} s$ 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
    $n^3+n^2+1$
  • B
    $n^3-n^2+1$
  • Correct answer
    C
    $n^3-n^2$
  • D
    $n^3+n^2$

Solution

Given, $a_0=1$ and $a_{n+1}=3 n^2+n+a_n$ $\Rightarrow a_n=3(n-1)^2+(n-1)+a_{n-1}$ Now, put $n=1,2,3, \ldots, n$ We get, $a_1=0+a_0=0+1=1$ $\begin{array}{ll}\Rightarrow & a_2=4+a_1 \\ \Rightarrow & a_3=14+a_2 \\ \Rightarrow & a_4=30+a_3\end{array}$ ........ ........ ........ $\begin{array}{ll}\Rightarrow & a_{n-1}=3(n-2)^2+(n-2)+a_{n-2} \\ \Rightarrow & a_n=3(n-1)^2+(n-1)+a_{n-1}\end{array}$ On adding, we get $a_n=\left(1+4+14+30+\ldots+3(n-1)^2+(n-1)\right)$ $\begin{array}{ll}\Rightarrow & a_n=\Sigma 3\left(n^2+1-2 n\right)+(n-1) \\ \Rightarrow & a_n=\Sigma\left(3 n^2+3-6 n+n-1\right) \\ \Rightarrow & a_n=\Sigma\left(3 n^2-5 n+2\right) \\ \Rightarrow & a_n=3 \Sigma n^2-5 \Sigma+2 \Sigma 1\end{array}$ $\Rightarrow a_n=\frac{3 n(n+1)(2 n+1)}{6}-\frac{5 n(n+1)}{2}+2 n$ $\Rightarrow a_n=\frac{n}{2}(n+1)\{2 n+1-5\}+2 n$ $\Rightarrow a_n=\frac{n}{2}(n+1)(2 n-4)+2 n$ $=n(n+1)(n-2)+2 n$ $\Rightarrow a_n=\left(n^2+n\right)(n-2)+2 n$ $=n^3-2 n^2+n^2-2 n+2 n$ $\Rightarrow a_n=n^3-n^2 (\because n \geq 0)$

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

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