Use mathematical induction to prove the truth of each of the following assertions for all n ≥1.
n³ + 5n is divisible by 6
I really do not understand this to much. This is what I have so far:
n = 1, 1³ - 5(1) = 6, which is divisible by 6
Then I really don't know how to go on from here. I appreciate any helpful replies. Thank you!
The next step is to assume the proposition is true for n.
The task is to show that if the proposition is true for n, then it would be true for n+1. Once that is established, then the proof is complete.
6|n^3+5n => 6|1+5 is true for n=1
Assume 6|N^3+5 is true for n, then
for n+1
(n+1)^3+5(n+1)
=n^3+5n +3n^2+3n+6
=n^3+5n +3n(n+1) + 6
We now examine the three terms:
n^3+5n is divisible by 6 by initial assumption.
6 is divisible by 6.
3n(n+1) falls into two cases:
1. n is odd, then n+1 is even, therefore 6 divides 3*(n+1)
2. n is even, then 6 divides 3n.
Since all three terms are divisible by 6, we only have to extract the factor of 6 from each term and declare the expression (n+1)^3+5(n+1) is also divisible by 6.
By the principle of mathematical induction, the proposition is proved. QED.
Thank you so much for your response! But I have completed that particular question. However, can you please help with this one? I am confused. . .
Use mathematical induction to establish the following formula.
n
Σ i² / [(2i-1)(2i+1)] = n(n+1) / 2(2n+1)
i=1
Thanks for any helpful replies :)
Any suggestions?
There are three steps:
1. Basis:
Test case for n=1 (or any other finite number):
n
Σ i² / [(2i-1)(2i+1)] = n(n+1) / 2(2n+1)
i=1
for n=1,
Left hand side=1/[(2*1-1)(2*1+2)=1/3
Right hand side=1(1+1)/[2(2*1+1)]=1/3
So formula is established for n=1.
2. Assume
formula is valid for case n.
3. Show that formula is valid for case n+1.
Left hand side:
n+1
Σ i² / [(2i-1)(2i+1)]
i=1
=
n
Σ i² / [(2i-1)(2i+1)]
i=1
+ (n+1)²/[(2*(n+1)-1)(2*(n+1)+2)]
=
n(n+1) / 2(2n+1) + (n+1)²/[(2*(n+1)-1)(2*(n+1)+2)]
=
[(n^2+n)*(2n+3)+2(n+1)^2] / [2(2n+1)(2n+3)]
=(n+1)(n+2)(2n+1) / [2(2n+1)(2n+3)]
=(n+1)((n+1)+1))/[2(2(n+1)+1)]
=m(m+1)/[2(2m+1)]
Which is precise the right hand side with m replacing n+1.
QED
To prove that the assertion "n³ + 5n is divisible by 6" is true for all n ≥ 1 using mathematical induction, you need to follow these steps:
Step 1: Base Case
Show that the statement is true for the base case, which is usually n = 1 in this case. You have already done that by substituting n = 1 into the expression:
1³ + 5(1) = 6, which is divisible by 6.
Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary positive integer k, where k ≥ 1. This is called the inductive hypothesis and is written as P(k), where P is the statement "n³ + 5n is divisible by 6".
Step 3: Inductive Step
Prove that the statement P(k+1) is also true, where P(k+1) is the statement "((k+1)³ + 5(k+1)) is divisible by 6".
Start with the left-hand side of P(k+1):
(k+1)³ + 5(k+1) = (k³ + 3k² + 3k + 1) + (5k + 5)
= k³ + 3k² + 3k + 5k + 6
Now, rearrange the terms:
= (k³ + 5k) + (3k² + 3k + 6)
= k(k² + 5) + 3(k² + k + 2)
From the inductive hypothesis, we know that k³ + 5k is divisible by 6. We also observe that k² + k + 2 is a positive integer for all positive integers k.
Since both k(k² + 5) and 3(k² + k + 2) are divisible by 6, their sum (k³ + 5k + 3k² + 3k + 6) is also divisible by 6.
Step 4: Conclusion
Since we have shown that P(k) implies P(k+1), we can conclude that the statement "n³ + 5n is divisible by 6" is true for all n ≥ 1 by mathematical induction.
Note: Induction is a rigorous mathematical proof technique, but it can be challenging to understand and apply at first. Take your time to understand each step, and feel free to ask for further clarification!