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Consider the set N = {2, 3, 4, ..., 2n + 1}, where n is a positive integer larger than 10005. If O is the average of the odd integers in N and £ is the average of the even integers in N then what is the value of O - E?
- a)0
- b)1
- c)n + 1/2n
- d)10006
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider the set N = {2, 3, 4, ..., 2n + 1}, where n is a positive int...
E=(2 + 4 + 6 + 8 + ... + 2 n)/n
O = (3 + 5 + 7 + 9 + ... + (2n + 1))/n
= [(2 + 1) + (4 + 1) + (6 + 1) + (8 + 1) + ... + (2 n + 1 )]/n
= (2 + 4 + 6 + 8 + . . . + 2n)ln + (1 + 1 + 1 + 1 + ... n times)/n
= E + 1
O - E = 1
Hence, option 2.
Note: The information that 'n is a positive integer larger than 10005' does not affect the answer in any way.
O = (3 + 5 + 7 + 9 + ... + (2n + 1))/n
= [(2 + 1) + (4 + 1) + (6 + 1) + (8 + 1) + ... + (2 n + 1 )]/n
= (2 + 4 + 6 + 8 + . . . + 2n)ln + (1 + 1 + 1 + 1 + ... n times)/n
= E + 1
O - E = 1
Hence, option 2.
Note: The information that 'n is a positive integer larger than 10005' does not affect the answer in any way.
Most Upvoted Answer
Consider the set N = {2, 3, 4, ..., 2n + 1}, where n is a positive int...
Solution:
Given, the set N = {2, 3, 4, ..., 2n - 1}, where n is a positive integer larger than 10005.
Let's find the sum of all the numbers in set N.
Sum of N = 2 + 3 + 4 + ... + (2n - 1)
= [(2 + 2n - 1) + (3 + 2n - 2) + (4 + 2n - 3) + ... + (n + n + 1)]
= n(2n + 1)/2
Now, let's find the sum of all the odd numbers in set N.
Sum of odd numbers = 3 + 5 + 7 + ... + (2n - 1)
= [(2n + 1) + (2n - 1) + (2n - 3) + ... + 3]
= n^2
As we know, the average of odd numbers in N is O.
So, O = Sum of odd numbers / Total number of odd numbers
= n^2 / n
= n
Now, let's find the sum of all the even numbers in set N.
Sum of even numbers = 2 + 4 + 6 + ... + (2n - 2)
= 2(1 + 2 + 3 + ... + (n - 1))
= n(n - 1)
As we know, the average of even numbers in N is E.
So, E = Sum of even numbers / Total number of even numbers
= n(n - 1) / n
= n - 1
Now, let's find the value of O - E.
O - E = n - (n - 1)
= 1
Therefore, the correct option is (b) 1.
Given, the set N = {2, 3, 4, ..., 2n - 1}, where n is a positive integer larger than 10005.
Let's find the sum of all the numbers in set N.
Sum of N = 2 + 3 + 4 + ... + (2n - 1)
= [(2 + 2n - 1) + (3 + 2n - 2) + (4 + 2n - 3) + ... + (n + n + 1)]
= n(2n + 1)/2
Now, let's find the sum of all the odd numbers in set N.
Sum of odd numbers = 3 + 5 + 7 + ... + (2n - 1)
= [(2n + 1) + (2n - 1) + (2n - 3) + ... + 3]
= n^2
As we know, the average of odd numbers in N is O.
So, O = Sum of odd numbers / Total number of odd numbers
= n^2 / n
= n
Now, let's find the sum of all the even numbers in set N.
Sum of even numbers = 2 + 4 + 6 + ... + (2n - 2)
= 2(1 + 2 + 3 + ... + (n - 1))
= n(n - 1)
As we know, the average of even numbers in N is E.
So, E = Sum of even numbers / Total number of even numbers
= n(n - 1) / n
= n - 1
Now, let's find the value of O - E.
O - E = n - (n - 1)
= 1
Therefore, the correct option is (b) 1.
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Community Answer
Consider the set N = {2, 3, 4, ..., 2n + 1}, where n is a positive int...
1
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Consider the set N = {2, 3, 4, ..., 2n + 1}, where n is a positive integer larger than 10005. If O is the average of the odd integers in N and is the average of the even integers in N then what is the value of O - E?a)0b)1c)n + 1/2nd)10006Correct answer is option 'B'. Can you explain this answer?
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