What is the first natural number that causes the statement 1 + 3 + 5 +... + (2n − 1) 4n − 1 to fail?
2
3
4
5
6
I think you lost an = sign somewhere.
thats how it is written
To find the first natural number that causes the statement to fail, we need to determine when the sum of the series 1 + 3 + 5 + ... + (2n - 1) is not equal to 4n - 1.
First, let's find the sum of the series using the formula for the sum of an arithmetic series:
S = (n/2)(first term + last term)
In the given series, the first term is 1 and the common difference between consecutive terms is 2. The last term is (2n - 1). Plugging these values into the formula, we get:
S = (n/2)(1 + 2n - 1)
= (n/2)(2n)
= n^2
Next, we need to find when the sum of the series is not equal to 4n - 1. So, we have:
n^2 ≠ 4n - 1
Rearranging the equation, we get a quadratic equation:
n^2 - 4n + 1 ≠ 0
To solve this equation, we can use the quadratic formula:
n = (-b ± sqrt(b^2 - 4ac)) / 2a
Here, a = 1, b = -4, and c = 1. Plugging these values into the quadratic formula, we get:
n = (4 ± sqrt((-4)^2 - 4(1)(1))) / 2(1)
= (4 ± sqrt(16 - 4)) / 2
= (4 ± sqrt(12)) / 2
Simplifying further, we have:
n = (4 ± 2sqrt(3)) / 2
= 2 ± sqrt(3)
Now, we need to find the first natural number for which this equation is not true. Since the series deals with natural numbers, we need to choose an integer value for n. If we round down (floor function) the solutions for n, we have n = 2 - sqrt(3) and n = 2 + sqrt(3).
Let's check the values of n = 2 and n = 3 in the equation to see if they cause the statement to fail:
For n = 2:
1 + 3 + 5 = 9 ≠ 7 = 4(2) - 1, so the statement fails.
For n = 3:
1 + 3 + 5 + 7 = 16 ≠ 11 = 4(3) - 1, so the statement fails.
Now, let's check the values of n = 4 and n = 5 to see if they cause the statement to fail:
For n = 4:
1 + 3 + 5 + 7 + 9 = 25 = 15 = 4(4) - 1, so the statement holds.
For n = 5:
1 + 3 + 5 + 7 + 9 + 11 = 36 ≠ 19 = 4(5) - 1, so the statement fails.
Therefore, the first natural number that causes the statement to fail is n = 5. Answer: 5