Q65 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
Using mathematical induction, the numbers $a_n \delta$ 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$n^3-n^2$
- B$n^3+n^2+1$
- Correct answerC$n^3-n^2+1$
- D$n^3+n^2$
Solution
The recurrence relation is given by $a_{n+1} = a_n + 3n^2 + n$ with $a_0 = 1$.
We can write the general term $a_n$ as a summation:
$a_n = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = 1 + \sum_{k=0}^{n-1} (3k^2 + k)$.
Using the standard summation formulas $\sum_{k=0}^{n-1} k^2 = \dfrac{(n-1)n(2n-1)}{6}$ and $\sum_{k=0}^{n-1} k = \dfrac{(n-1)n}{2}$, we substitute these into the expression for $a_n$:
$a_n = 1 + 3 \left[ \dfrac{(n-1)n(2n-1)}{6} \right] + \dfrac{(n-1)n}{2}$.
Simplifying the expression:
$a_n = 1 + \dfrac{n(n-1)(2n-1)}{2} + \dfrac{n(n-1)}{2} = 1 + \dfrac{n(n-1)}{2} [ (2n-1) + 1 ]$.
$a_n = 1 + \dfrac{n(n-1)}{2} [ 2n ] = 1 + n^2(n-1) = 1 + n^3 - n^2$.
Thus, $a_n = n^3 - n^2 + 1$.
Answer: $n^3-n^2+1$
We can write the general term $a_n$ as a summation:
$a_n = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = 1 + \sum_{k=0}^{n-1} (3k^2 + k)$.
Using the standard summation formulas $\sum_{k=0}^{n-1} k^2 = \dfrac{(n-1)n(2n-1)}{6}$ and $\sum_{k=0}^{n-1} k = \dfrac{(n-1)n}{2}$, we substitute these into the expression for $a_n$:
$a_n = 1 + 3 \left[ \dfrac{(n-1)n(2n-1)}{6} \right] + \dfrac{(n-1)n}{2}$.
Simplifying the expression:
$a_n = 1 + \dfrac{n(n-1)(2n-1)}{2} + \dfrac{n(n-1)}{2} = 1 + \dfrac{n(n-1)}{2} [ (2n-1) + 1 ]$.
$a_n = 1 + \dfrac{n(n-1)}{2} [ 2n ] = 1 + n^2(n-1) = 1 + n^3 - n^2$.
Thus, $a_n = n^3 - n^2 + 1$.
Answer: $n^3-n^2+1$
Hence, the correct answer is option: 'C'.