Answer :
Sure! Let's address each question step by step.
### Prove by Induction: [tex]\( n^2 + n \)[/tex] is even where [tex]\( n \)[/tex] is a positive integer
To prove that [tex]\( n^2 + n \)[/tex] is even for all positive integers [tex]\( n \)[/tex], we will use mathematical induction.
#### Base Case:
For [tex]\( n = 1 \)[/tex]:
[tex]\[ n^2 + n = 1^2 + 1 = 1 + 1 = 2 \][/tex]
2 is even. Therefore, the base case holds.
#### Inductive Step:
Assume that [tex]\( k^2 + k \)[/tex] is even for some positive integer [tex]\( k \)[/tex]. This is our inductive hypothesis.
We need to prove that [tex]\( (k+1)^2 + (k+1) \)[/tex] is even.
Consider:
[tex]\[ (k+1)^2 + (k+1) \][/tex]
[tex]\[ = k^2 + 2k + 1 + k + 1 \][/tex]
[tex]\[ = k^2 + 3k + 2 \][/tex]
Since we assumed [tex]\( k^2 + k \)[/tex] is even, we can write:
[tex]\[ k^2 + k = 2m \text{ for some integer } m \][/tex]
Now, consider:
[tex]\[ k^2 + 3k + 2 \][/tex]
[tex]\[ = (k^2 + k) + 2k + 2 \][/tex]
[tex]\[ = 2m + 2k + 2 \][/tex]
[tex]\[ = 2(m + k + 1) \][/tex]
Since [tex]\( m + k + 1 \)[/tex] is an integer, we have that [tex]\( 2(m + k + 1) \)[/tex] is even.
Therefore, [tex]\( (k+1)^2 + (k+1) \)[/tex] is even.
Since both the base case and inductive step have been proven, by mathematical induction, [tex]\( n^2 + n \)[/tex] is even for all positive integers [tex]\( n \)[/tex].
---
### Probability that Both Balls Drawn are Either White or Red
We are given 5 white balls, 4 red balls, and 3 blue balls in a box. We need to determine the probability that both balls drawn at random are either white or red.
#### Total Number of Balls:
[tex]\[ 5 + 4 + 3 = 12 \][/tex]
#### Total Number of Ways to Choose 2 Balls from 12:
[tex]\[ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \][/tex]
#### Number of Favorable Outcomes (both balls are either white or red):
We have 5 white balls and 4 red balls, making a total of 9 balls that are either white or red.
Number of ways to choose 2 balls from these 9 balls:
[tex]\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \][/tex]
#### Probability:
[tex]\[ P(\text{both balls are either white or red}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{36}{66} = \frac{6}{11} \][/tex]
So, the probability that both balls drawn are either white or red is [tex]\( \frac{6}{11} \)[/tex].
---
### Invite One or More of Your Five Friends to Your Birthday Party
We need to find the number of ways to invite one or more of your five friends to a birthday party.
#### Total Number of Friends:
5
Each friend can either be invited or not be invited. Therefore, for 5 friends, there are:
[tex]\[ 2^5 = 32 \text{ ways} \][/tex]
However, this includes the scenario where no friends are invited (1 way). To find the number of ways to invite one or more friends, we subtract the scenario where no friends are invited:
[tex]\[ 32 - 1 = 31 \][/tex]
So, there are 31 ways to invite one or more of your five friends to your birthday party.
### Prove by Induction: [tex]\( n^2 + n \)[/tex] is even where [tex]\( n \)[/tex] is a positive integer
To prove that [tex]\( n^2 + n \)[/tex] is even for all positive integers [tex]\( n \)[/tex], we will use mathematical induction.
#### Base Case:
For [tex]\( n = 1 \)[/tex]:
[tex]\[ n^2 + n = 1^2 + 1 = 1 + 1 = 2 \][/tex]
2 is even. Therefore, the base case holds.
#### Inductive Step:
Assume that [tex]\( k^2 + k \)[/tex] is even for some positive integer [tex]\( k \)[/tex]. This is our inductive hypothesis.
We need to prove that [tex]\( (k+1)^2 + (k+1) \)[/tex] is even.
Consider:
[tex]\[ (k+1)^2 + (k+1) \][/tex]
[tex]\[ = k^2 + 2k + 1 + k + 1 \][/tex]
[tex]\[ = k^2 + 3k + 2 \][/tex]
Since we assumed [tex]\( k^2 + k \)[/tex] is even, we can write:
[tex]\[ k^2 + k = 2m \text{ for some integer } m \][/tex]
Now, consider:
[tex]\[ k^2 + 3k + 2 \][/tex]
[tex]\[ = (k^2 + k) + 2k + 2 \][/tex]
[tex]\[ = 2m + 2k + 2 \][/tex]
[tex]\[ = 2(m + k + 1) \][/tex]
Since [tex]\( m + k + 1 \)[/tex] is an integer, we have that [tex]\( 2(m + k + 1) \)[/tex] is even.
Therefore, [tex]\( (k+1)^2 + (k+1) \)[/tex] is even.
Since both the base case and inductive step have been proven, by mathematical induction, [tex]\( n^2 + n \)[/tex] is even for all positive integers [tex]\( n \)[/tex].
---
### Probability that Both Balls Drawn are Either White or Red
We are given 5 white balls, 4 red balls, and 3 blue balls in a box. We need to determine the probability that both balls drawn at random are either white or red.
#### Total Number of Balls:
[tex]\[ 5 + 4 + 3 = 12 \][/tex]
#### Total Number of Ways to Choose 2 Balls from 12:
[tex]\[ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \][/tex]
#### Number of Favorable Outcomes (both balls are either white or red):
We have 5 white balls and 4 red balls, making a total of 9 balls that are either white or red.
Number of ways to choose 2 balls from these 9 balls:
[tex]\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \][/tex]
#### Probability:
[tex]\[ P(\text{both balls are either white or red}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{36}{66} = \frac{6}{11} \][/tex]
So, the probability that both balls drawn are either white or red is [tex]\( \frac{6}{11} \)[/tex].
---
### Invite One or More of Your Five Friends to Your Birthday Party
We need to find the number of ways to invite one or more of your five friends to a birthday party.
#### Total Number of Friends:
5
Each friend can either be invited or not be invited. Therefore, for 5 friends, there are:
[tex]\[ 2^5 = 32 \text{ ways} \][/tex]
However, this includes the scenario where no friends are invited (1 way). To find the number of ways to invite one or more friends, we subtract the scenario where no friends are invited:
[tex]\[ 32 - 1 = 31 \][/tex]
So, there are 31 ways to invite one or more of your five friends to your birthday party.
