n(n+1)(n+5)is a multiple of 3 proove it by mathematical induction.
Yipeeee!
This is a two step process.
1) show that by subbing in n=1 that the answer is a multiple of 3
2) assume true for n=k and show true for n= k + 1
n(n^2 + 6 n + 5)
n^3 + 6 n^2 + 5 n
if n = 3 k where k is any old integer
27 k^3 + 54 k^2 + 15 k
= 3 (9 k^3 + 18 k^2 + 5 k)
To prove that n(n+1)(n+5) is a multiple of 3 using mathematical induction, we need to follow these steps:
Step 1: Base Case
Start by proving the statement for the smallest possible value of n, which is typically n = 0 or n = 1.
Let's test the formula for n=0:
n(n+1)(n+5) = 0(0+1)(0+5) = 0(1)(5) = 0
0 is a multiple of 3 since 3 * 0 = 0.
Therefore, the base case is true.
Step 2: Inductive Hypothesis
Next, assume that the formula is true for an arbitrary value k, where k represents any non-negative integer.
For k = 0, the formula is true.
Step 3: Inductive Step
Now, we need to prove that if the formula is true for k, then it is also true for k + 1.
Assume that n(n+1)(n+5) is divisible by 3 for some arbitrary value k.
We want to show that (k+1)[(k+1)+1][(k+1)+5] is also divisible by 3.
Expanding the expression, we get:
(k+1)(k+2)(k+6)
(k^2+3k+2)(k+6)
k^3 + 9k^2 + 20k + 12
Now, let's express this in terms of k:
k^3 + 9k^2 + 20k + 12 = (k^3 + 3k^2 + 3k) + (6k^2 + 17k + 12)
= k(k^2 + 3k + 3) + (2k + 3)(3k + 4)
By the inductive hypothesis, we know that k(k^2 + 3k + 3) is divisible by 3.
We also notice that (2k + 3)(3k + 4) is a multiple of 3 since both terms are divisible by 3 (either 2k + 3 or 3k + 4).
Therefore, (k+1)[(k+1)+1][(k+1)+5] is divisible by 3.
Step 4: Conclusion
By following the steps of mathematical induction, we have shown that if the formula is true for k, then it is also true for k + 1. The base case was established, so we can conclude that the statement "n(n+1)(n+5) is a multiple of 3" holds for all non-negative integers n.