Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.

प्रश्न

Let P(n): “2ⁿ < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.

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MCQ

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उत्तर

Let P(n): “2ⁿ < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is 4.

Explanation:

P(1): 2 < 1 is false.

P(2): 2² < 1 × 2 is false.

P(3): 2³ < 1 × 2 × 3 is false.

But P(4): 2⁴ < 1 × 2 × 3 × 4 is true.

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Principle of Mathematical Induction

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 11 Q 10 Q 12

पाठ 4: Principle of Mathematical Induction - Solved Examples [पृष्ठ ६९]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11

पाठ 4 Principle of Mathematical Induction
Solved Examples | Q 11 | पृष्ठ ६९

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Principle of Mathematical Induction

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संबंधित प्रश्‍न

Prove the following by using the principle of mathematical induction for all n ∈ N:

1.3 + 2.3^3 + 3.3^3 +...+ n.3^n = `((2n -1)3^(n+1) + 3)/4`

Prove the following by using the principle of mathematical induction for all n ∈ N:

1.3 + 3.5 + 5.7 + ...+(2n -1)(2n + 1) = `(n(4n^2 + 6n -1))/3`

Prove the following by using the principle of mathematical induction for all n ∈ N: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^n`

Prove the following by using the principle of mathematical induction for all n ∈ N:

`(1+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 1)`

Prove the following by using the principle of mathematical induction for all n ∈ N:

`1^2 + 3^2 + 5^2 + ... + (2n -1)^2 = (n(2n - 1) (2n + 1))/3`

Prove the following by using the principle of mathematical induction for all n ∈ N:

`1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/((2n + 1)(2n +3)) = n/(3(2n +3))`

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Prove the following by using the principle of mathematical induction for all n ∈ N: x²ⁿ – y²ⁿ is divisible by x + y.

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Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.

1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .

1 + 3 + 3² + ... + 3ⁿ⁻¹ = \[\frac{3^n - 1}{2}\]

\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

1² + 3² + 5² + ... + (2n − 1)² = \[\frac{1}{3}n(4 n^2 - 1)\]

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(ab)ⁿ = aⁿbⁿ for all n ∈ N.

2.7ⁿ + 3.5ⁿ − 5 is divisible by 24 for all n ∈ N.

11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133 for all n ∈ N.

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x²ⁿ⁻¹ + y²ⁿ⁻¹ is divisible by x + y for all n ∈ N.

\[\text{ Prove that } \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all n } \in N .\]

\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
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\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]

\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]

\[\text { A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that } x_n = \frac{2}{n!} \text{ for all } n \in N .\]

Prove by method of induction, for all n ∈ N:

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Answer the following:

Prove, by method of induction, for all n ∈ N

`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`

Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

1 + 3 + 5 + ... + (2n – 1) = n²