I eliminated 2 of the 5 and now I have 3 I am confused towards which is the right one? help mathematical induction to determine which formula is true for all natural numbers n ≥ 3
a) 2n2 > (n + 1)2
b) (n − 1)2 > n2 − 6
c) (n + 1)2 < n2 +
a. Let n = 3.
2*3^2 > (3+1)^2.
18 > 16. True.
Let n = 4.
2*4^2 > (4+1)^2.
32 > 25. True.
Check b, and c.
(n+1)^2 = n^2+2n+1
So, check to see whether
2n^2 > n^2+2n+1
n^2 > 2n+1
Clearly, if n >= 3, n^2 >= 3n > 2n+1
(n-1)^2 > n^2 - 6
n^2 - 2n + 1 > n^2 - 6
2n - 1 < 6
2n < 7
Not true for n > 3
(c) is incomplete, but should be easy to check in similar ways.
To determine which formula is true for all natural numbers n ≥ 3, we can use mathematical induction. Mathematical induction is a proof technique used to establish statements that hold for all natural numbers.
To begin, we will assume that one of the formulas, let's say formula a) 2n^2 > (n + 1)^2, is true for some natural number n = k. This is called the base case.
Now, we need to show that if the formula is true for some natural number k, it must also be true for the next natural number k + 1. This is called the induction step.
1. Base case (n = k):
Let's substitute n = k into formula a): 2k^2 > (k + 1)^2
Now, simplify both sides of the equation:
2k^2 > (k^2 + 2k + 1)
2. Induction step (n = k + 1):
Assuming the formula is true for n = k, we now substitute n = k + 1 into formula a): 2(k + 1)^2 > ((k + 1) + 1)^2
Now, simplify both sides of the equation:
2(k^2 + 2k + 1) > ((k^2 + 2k + 1) + 2(k + 1) + 1)
2k^2 + 4k + 2 > k^2 + 4k + 3
Now, let's combine like terms:
k^2 + 2k + 2 > k^2 + 4k + 3
Next, subtract k^2 from both sides:
2k + 2 > 4k + 3 - k^2
Subtract 3 from both sides:
2k - 1 > 4k - k^2
Rearrange the terms:
k^2 - 2k + 1 > 0
Now, simplify the left side of the inequality:
(k - 1)^2 > 0
Since (k - 1)^2 is always positive for any natural number k, this inequality is true for all n = k + 1.
Therefore, we have shown that if the formula is true for some natural number k, it will also be true for the next natural number k + 1.
Finally, since we have shown that the formula is true for n = k (base case) and that if it is true for n = k, it will be true for n = k + 1 (induction step), by mathematical induction, we can conclude that formula a) 2n^2 > (n + 1)^2 is true for all natural numbers n ≥ 3.
You can repeat the above process for each of the given formulas, substituting n = k and n = k + 1 into each formula, to determine which one holds true for all natural numbers n ≥ 3.