Q. Using mathematical induction, then numbers $a_{n}'s$ are defined by $a_{0}$=1, $a_{n + 1} = {3n}^{2} + n + a_{n},(n \geq 0)$ Then, $a_{n}$ is equal to
A) $n^{3} + n^{2} + 1$
B) $n^{3} - n^{2} + 1$
C) $n^{3} - n^{2}$
D) $n^{3} + n^{2}$
Explanation
(b)
Given, $a_{0} = 1,\ a_{n + 1} = {3n}^{2} + n + a_{n}$
$\Longrightarrow a_{1} = 3(0) + 0 + a_{0} = 1$
$\Longrightarrow a_{2} = 3({1)}^{2} + 1 + a_{1} = 3 + 1 + 1 = 5$
From option (b),
Let $P(n) = n^{3} - n^{2} + 1$
$\therefore\ \ \ P(0) = 0 - 0 + 1 = 1 = a_{0}$
$\ \ \ \ \ \ P(1) = 1^{3} - 1^{2} + 1 = 1 = a_{1}$
$and\ \ P(2) = {(2)}^{3} - {(2)}^{2} + 1 = 5 = a_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.