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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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If 4a - 13 = 6b + 35 = 8c -17 = d, find the smallest possible value of a + b + c + d.
Afina-wow [57]

Answer:

The smallest possible value of a+b+c+d is: 4

Step-by-step explanation:

since we are given that:

4a - 13 = 6b + 35 = 8c -17 = d

<em>on taking the first two equality i.e. 4a-13=6b+35</em>

we get b=\dfrac{2}{3}a-8

<em>on using the first and third equality we have:</em>

4a-13=8c-17

c=\dfrac{1}{2}a+\dfrac{1}{2}

<em>also from the first and last equality we have:</em>

d=4a-13

Hence,

a+b+c+d=a+\dfrac{2}{3}a-8+\dfrac{1}{2}a+\dfrac{1}{2}+4a-13\\\\a+b+c+d=\dfrac{37a}{6}-\dfrac{41}{2}

a+b+c+d=\dfrac{37a-123}{6}

the smallest possible value such that the expression a+b+c+d is positive will be claculated as:

a+b+c+d>0

that means \dfrac{37a-123}{6}>0

a>3.324

But as a is an integer, hence the smallest such value is 4.


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