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natulia [17]
3 years ago
13

Question 31 pts Prove the statement is true using mathematical induction: 2n-1 ≤ n! Use the space below to write your answer. To

make the < symbol, you might want to use the < with the underline feature.
Mathematics
1 answer:
Sladkaya [172]3 years ago
4 0

Answer:

P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Step-by-step explanation:

Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3

Basis: P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:

2(k+1) - 1 = 2k + 2 - 1

≤ 2 + k! (by the inductive hypothesis)

= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.

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Resuelve las siguientes ecuaciones: -x^2 - x =0 -X^2 + 12x = 0 3X^2 - 9x = 0 X^2 - 2x - 3 = 0
vaieri [72.5K]

Responder:

<em>a) 1 </em>

<em>b) 12 </em>

<em>c) 3 </em>

<em>d) -1 y 3 </em>

Explicación paso a paso:

Dadas las siguientes ecuaciones;

a) x²-x =0

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b) -x²+12x = 0

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Suma 9x a ambos lados

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3x = 9

x = 9/3

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d) x²-2x-3 = 0

x²-3x+x-3 = 0

x(x-3)+1(x-3) = 0

(x+1)(x-3) = 0

x+1 = 0 y x-3 = 0

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