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)