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Standard XI

Mathematics

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 $a_{n} =$

Q. 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} =

Q. The sequence

(a_{n}) n \geq 0

is defined by

a_{0} = a_{1} = 1

and

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

for all

n \geq 1.

Prove that the number

2 a_{n} - 1

is a square for all

n \geq 0.

Q. Given

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

and

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

Find

n \geq 2

where a > 0, A > 0, prove that

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

using mathematical induction.

Q. Let S be the infinite sum given by

S = \infty \sum n = 0 \frac{a_{n}}{10^{2 n}}

, where

(a_{n})_{n \geq 0}

is a sequence defined by

a_{0} = a_{1} = 1

and

a_{j} = 20 a_{j - 1}

for

j \geq 2

. If

S

is expressed in the form

\frac{a}{b}

, where

a, b

are coprime positive integers, than

a

equals.

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Using mathematical Induction, the numbers an ′s are defined by a0=1, an+1=3n2+n+an(n≥0) Then an=

The correct option is B n3−n2+1 Putting n = 0, 1, 2, ………………. n – 1 and adding an−a0=3∑(n−1)2+∑n⇒an=n3−n2+1

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 $a_{n} =$

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

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