Use mathematical induction to prove that for all integers n ≥ 5, 1 + 4n < 2n
To prove the inequality 1 + 4n < 2n for all integers n ≥ 5 using mathematical induction, we need to follow these steps:
Step 1: Base Case
For the base case, we need to show that the inequality holds true for n = 5.
Substituting n = 5 into the inequality, we get:
1 + 4(5) < 2(5)
1 + 20 < 10
21 < 10
Since the inequality is false for n = 5, we cannot proceed with the proof by mathematical induction. Therefore, the statement "1 + 4n < 2n" is not true for all integers n ≥ 5.
To prove a statement using mathematical induction, we typically follow these steps:
Step 1: Base Case. We start by showing that the statement is true for the smallest value of n. In this case, n ≥ 5, so we need to prove the statement for n = 5.
Step 2: Inductive Hypothesis. Assume that the statement is true for some arbitrary value k ≥ 5. This is called the inductive hypothesis.
Step 3: Inductive Step. Using the inductive hypothesis, we need to show that the statement is also true for the next value, which is k + 1.
Let's follow these steps to prove the statement using mathematical induction:
Step 1: Base Case
For n = 5:
1 + 4(5) = 1 + 20 = 21
2(5) = 10
So, the statement is true for n = 5.
Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary value k ≥ 5:
1 + 4k < 2k
Step 3: Inductive Step
We need to prove that the statement is true for k + 1:
1 + 4(k + 1) < 2(k + 1)
To start the inductive step, let's simplify the left side of the inequality:
1 + 4(k + 1) = 1 + 4k + 4 = (1 + 4k) + 4
Using the inductive hypothesis:
(1 + 4k) + 4 < 2k + 4
We can see that 2k + 4 is equal to 2(k + 1), so the inequality becomes:
(1 + 4k) + 4 < 2(k + 1)
Simplifying further:
(1 + 4k) + 4 = 1 + 4k + 4 = 5 + 4k
Now, we need to show that 5 + 4k < 2(k + 1).
Let's simplify the right side of the inequality:
2(k + 1) = 2k + 2
Using the assumption that k ≥ 5:
2k + 2 > 4k
Now, we can see that 5 + 4k < 2k + 2 < 2k + 4, which means:
5 + 4k < 2(k + 1)
Therefore, the statement is true for k + 1.
Since the statement is true for the base case (n = 5), and we have shown that if it is true for some value k, then it is also true for k + 1, we can conclude that the statement is true for all integers n ≥ 5.
Thus, using mathematical induction, we have proved that for all integers n ≥ 5, 1 + 4n < 2n.
try n=5
1+4*5 < 2^5 -- true
So, what about n+1?
1+4(n+1) = 1+4n + 4
Now just show that it is less than 2^(n+1)