Proofs of Mathematical Induction: Examples + Steps

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In summary, using mathematical induction, we can prove the following statements: 1. -1/2, -1/4, -1/8... -1/2^n = (1/(2^n))-12. a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]3. 1^3 + 2^3 + 3^3... + n^3 = (n^2(n+1)^2)/(4)4. For all positive integral values of n, (3^(4n))-1 is divisible by 80. The proof for each statement involves verifying the formula for
  • #1
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"Prove each of the following using Mathematical induction;" show all steps
pleasee someone help.. i have a test on this tommorow and i just need some examples.. could you also try to show all steps including the words so i understand how you got there? thakns so much in advance~!

1. -1/2, -1/4, -1/8... -1/2^n = (1/(2^n))-1

2. a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]

3. 1^3 + 2^3 + 3^3... + n^3 = (n^2(n+1)^2)/(4)

4. show that (3^(4n))-1 is dividislbe by 80 for all positive integral values of n
 
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  • #2
Let's take 3.
Part 1: Verification of the formula for n=1:
We have:
[tex]\frac{1^{2}(1+1)^{2}}{4}=\frac{1*4}{4}=1=1^{3}[/tex]
That is, the formula is true for n=1.
Part 2: The induction step
Suppose it is true for n=k-1:
Then, we have:
[tex]1^{3}+2^{3}+++(k-1)^{3}+k^{3}=\frac{(k-1)^{2}k^{2}}{4}+k^{3}[/tex]
since the formula holds for n=k-1.
Furthermore, we have:
[tex]\frac{(k-1)^{2}k^{2}}{4}+k^{3}=\frac{(k-1)^{2}k^{2}+4k^{3}}{4}=\frac{k^{2}((k-1)^{2}+4k)}{4}=\frac{k^{2}(k^{2}+2k+1)}{4}=\frac{k^{2}(k+1)^{2}}{4}[/tex]
But that is precisely what the formula would predict it to be!

Thus, the formula is correct.
 
  • #3
thanks! okay i did the same thing for number one and i got

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

how do i cancel this out?
 
  • #4
Lowest common denominator then separate 2^(k+1)
 
  • #5
like 1/2^k + 1/2^k * 1/2 = 1/2^k - 2k/2k - (1/2^k * 1/2)?
where do i go from there?
 

What is mathematical induction?

Mathematical induction is a method of mathematical proof used to prove statements about natural numbers or other well-ordered sets. It is based on the principle that if a statement is true for a certain number, and if it is also true for the next number, then it is true for all subsequent numbers.

What is the purpose of using mathematical induction?

The purpose of using mathematical induction is to prove that a statement is true for all natural numbers, without having to explicitly check each individual number. It is a powerful tool for proving statements that involve infinite sets of numbers.

What are the steps involved in a proof by mathematical induction?

The steps involved in a proof by mathematical induction are:

  1. Base case: Prove that the statement is true for the first natural number (usually 0 or 1).
  2. Inductive hypothesis: Assume that the statement is true for an arbitrary but fixed natural number, usually denoted as k.
  3. Inductive step: Use the inductive hypothesis to prove that the statement is also true for the next natural number, k+1.
  4. Conclusion: By the principle of mathematical induction, the statement is true for all natural numbers.

What are some common mistakes to avoid when using mathematical induction?

Some common mistakes to avoid when using mathematical induction are:

  • Not proving the base case or assuming it is true without proof.
  • Using the wrong variable for the inductive hypothesis.
  • Not clearly stating the inductive hypothesis and inductive step.
  • Assuming that the statement is true for all natural numbers without explicitly proving it.
  • Not using the correct form of the statement being proved in the inductive step.

Can mathematical induction be used to prove statements about sets other than natural numbers?

Yes, mathematical induction can be used to prove statements about any well-ordered set. This includes sets such as integers, rational numbers, and even some infinite sets like the set of all positive real numbers. However, the base case and inductive step may need to be adjusted depending on the specific set being used.

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