What is the first natural number that causes the statement 1 + 3 + 5 +... + (2n − 1)<=4n − 1 to fail?
A) 2
B) 3
C) 4
D) 5
E) 6
start checking. How about showing some of your own ideas on further problems? I can do them - the goal is for you to work on them, eh?
Honestly, i don't get any of them..
it asks for the first natural number ...
the natural numbers are 1,2,3,...
So, start trying them:
n=1: (2n-1) = 1
1 <= 4*1-1
1 <= 3? yes
n=2: (2n-1) = 3
1+3 <= 4*2-1
4 <= 7? yes
n=3: (2n-1) = 5
1+3+5 <= 4*3-1
9 <= 11? yes
n=4: (2n-1) = 7
1+3+5+7 <= 4*4-1
16 <= 15? NO
Or, you can solve the inequality. As you can see,
1+3+5+...+(2n-1) = n^2
So, we want to see where
x^2 <= 4n-1
x^2 - 4n + 1 <= 0
this parabola has a root at 2+√3 = 3.7
So, as long as x <= 3.7, x^2 <= 4n-1
The first natural number greater than 3.7 is 4.
Thank you!! you really are helping me alot.
To find the first natural number that causes the given statement to fail, we need to find the smallest value of n that violates the inequality: 1 + 3 + 5 + ... + (2n − 1) <= 4n − 1.
First, let's simplify the left side of the inequality. The sum of the odd numbers from 1 to (2n − 1) can be expressed as the sum of an arithmetic series:
1 + 3 + 5 + ... + (2n − 1) = n^2
Now, let's compare this to the right side of the inequality (4n − 1):
n^2 <= 4n − 1
Rearranging the inequality, we have:
n^2 - 4n <= -1
To solve this quadratic inequality, we can set it equal to zero and find the values of n that satisfy it:
n^2 - 4n + 1 = 0
Using the quadratic formula, we can find the solutions to this equation:
n = (4 ± sqrt(16 - 4))/2
n = (4 ± sqrt(12))/2
n = 2 ± sqrt(3)
Since n must be a natural number, we take the positive value of the square root:
n = 2 + sqrt(3)
Checking the answer options, the smallest value of n that satisfies this condition is between 2 and 3. Therefore, the first natural number that causes the given statement to fail is n = 3.
Therefore, the correct answer is option B) 3.