Proof Using Induction: Discrete Maths Problem Solving

[Ryuna]

Can someone help me solve the following problems from Discrete Maths, using induction to prove them
View attachment 2554
Thanks in advance

 

[MarkFL]

Hello, and welcome to MHB! :D

In the future, we ask that you post no more than two questions in a thread and that you show what you have tired so we know where you are stuck, and can best help.

I am assuming the first one is instead:

\(\displaystyle \sum_{k=1}^n\left(3\cdot2^{k-1}\right)=3\left(2^n-1\right)\)

The first step is to demonstrate that the base case (for $n=1$) is true. Have you done that?

 

[Ryuna]

^Right, I'll keep that in mind.

And the case is true for n=1. These are practice questions for my exam and I was looking for key answers to use as indicators. I'll be back after trying more.

could you give me a hint on how to start with the last one?

 

[MarkFL]

...could you give me a hint on how to start with the last one?

Well, after showing the base case is true, you want to state your induction hypothesis. It appears to me that using the given recursion will be key to your induction step. I would write the recursion in terms of $n+1$, and see what can be done with that.

 

[blue_raver22]

Sure, I would be happy to assist you in solving these problems using induction.

Firstly, let's review what induction is. Induction is a mathematical proof technique used to prove statements about all natural numbers. It involves proving a base case, usually for n=1, and then showing that if the statement is true for n=k, it must also be true for n=k+1. This process is repeated until the desired statement is proven for all natural numbers.

Now, let's take a look at the problems from Discrete Maths that you need help with:

1. Prove that 1+3+5+...+(2n-1) = n^2 for all natural numbers n.

Solution:
Base case: For n=1, we have 1 = 1^2, which is true.

Inductive hypothesis: Assume that the statement is true for n=k, i.e. 1+3+5+...+(2k-1) = k^2.

Inductive step: Now, we need to show that if the statement is true for n=k, it must also be true for n=k+1. So, let's add (2k+1) to both sides of our inductive hypothesis:
1+3+5+...+(2k-1)+(2k+1) = k^2 + (2k+1)
= (k+1)^2
= 1+3+5+...+(2k-1)+(2k+1)
Therefore, the statement is true for n=k+1.

By the principle of mathematical induction, the statement is true for all natural numbers n.

2. Prove that 1^3 + 2^3 + 3^3 + ... + n^3 = (n(n+1)/2)^2 for all natural numbers n.

Solution:
Base case: For n=1, we have 1^3 = (1(1+1)/2)^2, which is true.

Inductive hypothesis: Assume that the statement is true for n=k, i.e. 1^3 + 2^3 + 3^3 + ... + k^3 = (k(k+1)/2)^2.

Inductive step: Now, we need to show that if the statement is true for n=k, it must also be true for n=k+1. So, let's add (