n^3+2n is the multiple of 3 prove it by math induction method
step 1:
test for n = 1
1^3 + 2(1) = 3 , which is a multiple of 3
step 2:
assume it is true for n = k
that is, k^3 + 3k is a multiple of 3, or it is divisible by 3
step 3:
show that it is also true for n = k+1
that is, show that (k+1)^3 + 3k is a multiple of 3
let's take the difference
(k+1)^3 + 2(k+1) - (k^3 + 2k)
= k^3 + 3k^2 + 3k + 1 + 2k + 2 - k^3 - 2k
= 3k^2 + 3k + 3
= 3(k^2 + k + 1)
which is divisible by 3, (since 3 is a factor)
so n^3 + 3n is always a multiple of 3
the property I used is the following:
if 2 numbers are divisible by the same number, then their difference is divisible by that same number
e.g. 91 and 49 are both divisible by 7
then 91-49 or 42 is also divisible by 7
-- try it for other numbers.
since we knew the second number, k^3 + 2k , was divisible by 3 and the result was divisible by 3, then the first number, (k+1)^3 + 2(k+1) has to be divisible by 3
To prove that n^3 + 2n is a multiple of 3 using the mathematical induction method, we need to follow these steps:
Step 1: Base Case
First, we establish the base case. We will start by verifying if the statement holds true for the smallest possible value of n, which is n = 1 in this case.
When n = 1:
n^3 + 2n = 1^3 + 2(1) = 1 + 2 = 3
As we can see, when n = 1, n^3 + 2n equals 3, which is a multiple of 3. Therefore, our base case is satisfied.
Step 2: Inductive Hypothesis
Next, we assume that the statement holds true for a specific value, let's call it k. This is known as the inductive hypothesis.
Assuming that n = k, which implies that n^3 + 2n is a multiple of 3:
n^3 + 2n = k^3 + 2k (Inductive Hypothesis)
Step 3: Inductive Step
We need to prove that the statement holds true for the next value, k + 1.
When n = k + 1:
(n + 1)^3 + 2(n + 1) = (k + 1)^3 + 2(k + 1) (Substituting n with k + 1)
Expanding the expression:
(k + 1)^3 + 2(k + 1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 (Expanding (k + 1)^3)
Combining like terms:
k^3 + 3k^2 + 3k + 1 + 2k + 2 = k^3 + 3k^2 + 5k + 3
Now, let's consider n^3 + 2n:
n^3 + 2n = k^3 + 2k (Inductive Hypothesis)
We can notice that k^3 + 3k^2 + 5k + 3 is equal to (k^3 + 2k) + (3k^2 + 3k + 3).
Since we have assumed that n^3 + 2n is a multiple of 3 for n = k, and we can represent k^3 + 3k^2 + 5k + 3 as (k^3 + 2k) + (3k^2 + 3k + 3), the next step is to prove that (3k^2 + 3k + 3) is also a multiple of 3.
Simplifying (3k^2 + 3k + 3):
3k^2 + 3k + 3 = 3(k^2 + k + 1)
As we can see, (3k^2 + 3k + 3) is a multiple of 3, since it can be expressed as the product of 3 and (k^2 + k + 1).
Hence, if n = k + 1, we can conclude that n^3 + 2n is also a multiple of 3.
Step 4: Conclusion
We have proven the base case and the inductive step, which completes the proof by mathematical induction. Therefore, it has been shown that for all positive integers n, n^3 + 2n is a multiple of 3.