Using PMI prove un=((-1)^(n+1))/3 + (2^n)/3

Thread starter AerospaceEng
Start date Aug 30, 2011

In summary, to use PMI (Principle of Mathematical Induction) to prove the sequence un=((-1)^(n+1))/3 + (2^n)/3, you need to follow these steps: 1. Show that the statement is true for n=1 (base case). 2. Assume that the statement is true for some arbitrary integer k (inductive hypothesis). 3. Use the inductive hypothesis to prove that the statement is also true for k+1 (inductive step). 4. By using PMI, conclude that the statement is true for all positive integers n. The base case for this proof is when n=1. The inductive hypothesis is to assume the statement is true for

Aug 30, 2011

AerospaceEng

Solved thank you.

Thank you for all the help just a little forgetful mistake.

Last edited: Aug 30, 2011

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Aug 30, 2011

micromass

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You didn't use the induction hypothesis yet. You need to sub

[tex]u_n = \frac{(-1)^{n+1}+2^n}{3}[/tex]

and

[tex]u_{n-1} = \frac{(-1)^n+2^{n-1}}{3}[/tex]

FAQ: Using PMI prove un=((-1)^(n+1))/3 + (2^n)/3

How do you use PMI to prove un=((-1)^(n+1))/3 + (2^n)/3?

To use PMI (Principle of Mathematical Induction) to prove the sequence un=((-1)^(n+1))/3 + (2^n)/3, we need to follow these steps:
1. First, show that the statement is true for n=1 (base case).
2. Then, assume that the statement is true for some arbitrary integer k (inductive hypothesis).
3. Next, use the inductive hypothesis to prove that the statement is also true for k+1 (inductive step).
4. Lastly, by using PMI, we can conclude that the statement is true for all positive integers n.

What is the base case when using PMI to prove un=((-1)^(n+1))/3 + (2^n)/3?

The base case when using PMI to prove un=((-1)^(n+1))/3 + (2^n)/3 is when n=1. This means that we need to show that the statement is true for n=1 before proceeding to the inductive step.

How do you write the inductive hypothesis when using PMI to prove un=((-1)^(n+1))/3 + (2^n)/3?

The inductive hypothesis for proving un=((-1)^(n+1))/3 + (2^n)/3 is to assume that the statement is true for some arbitrary integer k, which is represented as un=((-1)^k+1)/3 + (2^k)/3.

What is the inductive step when using PMI to prove un=((-1)^(n+1))/3 + (2^n)/3?

The inductive step when using PMI to prove un=((-1)^(n+1))/3 + (2^n)/3 is to use the inductive hypothesis (un=((-1)^k+1)/3 + (2^k)/3) to prove that the statement is also true for k+1. This usually involves algebraic manipulation and substitution of k+1 in place of k.

Why is it important to use PMI to prove un=((-1)^(n+1))/3 + (2^n)/3?

Using PMI to prove un=((-1)^(n+1))/3 + (2^n)/3 is important because it is a rigorous and systematic method of proving mathematical statements for all positive integers. It helps to eliminate the need for testing each individual case and provides a clear and logical proof for the statement. Additionally, it is a fundamental principle in mathematical induction, which is widely used in various fields of mathematics and science.