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Proof by mathematical induction

Using mathema...

Question

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 $a_{n} =$

$n^{3} + n^{2} + 1$

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$n^{3} - n^{2} + 1$

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$n^{3} - n^{2}$

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$n^{3} + n^{2}$

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Solution

The correct option is B
$n^{3} - n^{2} + 1$

Putting n = 0, 1, 2, ………………. n – 1 and adding $a_{n} - a_{0} = 3 \sum {(n - 1)}^{2} + \sum n \Rightarrow a_{n} = n^{3} - n^{2} + 1$

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Similar questions

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 $a_{n} =$

a_{n}^{'} s

a_{0} = 1, a_{n + 1} = 3 n^{2} + n + a_{n} (n \geq 0) .

a_{n} =

(a_{n}) n \geq 0

a_{0} = a_{1} = 1

a_{n + 1} = 14 a_{n} - a_{n - 1}

n \geq 1.

2 a_{n} - 1

n \geq 0.

a_{1} = \frac{1}{2} (a_{0} + \frac{A}{a_{0}}), a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}})

a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}})

n \geq 2

\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{(A)}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{(A)}})}^{2^{n - 1}}

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