Q101 / 180 · COMEDK 2021 · 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 answerB$n^{3}-n^{2}+1$
- C$n^{3}-n^{2}$
- D$n^{3}+n^{2}$
Solution
Given the recurrence relation $a_{n+1} = a_{n} + 3n^{2} + n$ with $a_{0} = 1$.
We can write the terms as:
$a_{n} = a_{0} + \sum_{k=0}^{n-1} (a_{k+1} - a_{k})$
$a_{n} = 1 + \sum_{k=0}^{n-1} (3k^{2} + k)$
Using the standard summation formulas $\sum_{k=0}^{n-1} k^{2} = \dfrac{(n-1)n(2n-1)}{6}$ and $\sum_{k=0}^{n-1} k = \dfrac{(n-1)n}{2}$:
$a_{n} = 1 + 3 \left[ \dfrac{(n-1)n(2n-1)}{6} \right] + \dfrac{(n-1)n}{2}$
$a_{n} = 1 + \dfrac{n(n-1)(2n-1)}{2} + \dfrac{n(n-1)}{2}$
$a_{n} = 1 + \dfrac{n(n-1)}{2} [ (2n-1) + 1 ]$
$a_{n} = 1 + \dfrac{n(n-1)(2n)}{2}$
$a_{n} = 1 + n^{2}(n-1)$
$a_{n} = 1 + n^{3} - n^{2}$
Answer: $n^{3}-n^{2}+1$
We can write the terms as:
$a_{n} = a_{0} + \sum_{k=0}^{n-1} (a_{k+1} - a_{k})$
$a_{n} = 1 + \sum_{k=0}^{n-1} (3k^{2} + k)$
Using the standard summation formulas $\sum_{k=0}^{n-1} k^{2} = \dfrac{(n-1)n(2n-1)}{6}$ and $\sum_{k=0}^{n-1} k = \dfrac{(n-1)n}{2}$:
$a_{n} = 1 + 3 \left[ \dfrac{(n-1)n(2n-1)}{6} \right] + \dfrac{(n-1)n}{2}$
$a_{n} = 1 + \dfrac{n(n-1)(2n-1)}{2} + \dfrac{n(n-1)}{2}$
$a_{n} = 1 + \dfrac{n(n-1)}{2} [ (2n-1) + 1 ]$
$a_{n} = 1 + \dfrac{n(n-1)(2n)}{2}$
$a_{n} = 1 + n^{2}(n-1)$
$a_{n} = 1 + n^{3} - n^{2}$
Answer: $n^{3}-n^{2}+1$
Hence, the correct answer is option: 'B'.