1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gnesinka [82]
3 years ago
14

Prove for any positive integer n, n^3 +11n is a multiple of 6

Mathematics
2 answers:
Margarita [4]3 years ago
7 0

Answer:

Prove by induction that n3+11n is divisible by 6 for every positive integer n.

I've started by letting P(n)=n3+11n

P(1)=12 (divisible by 6, so P(1) is true.)

Assume P(k)=k3+11k is divisible by 6.

P(k+1)=(k+1)3+11(k+1)=k3+3k2+3k+1+11k+11=(k3+11k)+(3k2+3k+12)

Since P(k) is true, (k3+11k) is divisible by 6 but I can't show that (3k2+3k+12) is divisible by 6

Step-by-step explanation:

suter [353]3 years ago
6 0

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

You might be interested in
On the Jackals baseball team, there are 24 right-handed players and 6 left-handed players. What is the ratio of right-handed pla
mamaluj [8]

Answer:

4:1

24 divided by 6 is 4.

Every time you add 1 left handed, you add 4 right handed.

3 0
3 years ago
Perform the indicated operations. Do the root of -48 in terms of i please
lubasha [3.4K]

Answer:

\frac{-5-\sqrt{-48}}{40}  in terms of i is: \frac{-1}{8}-\frac{i\,\sqrt{3}}{10}

Step-by-step explanation:

\frac{-5-\sqrt{-48}}{40}

We know that \sqrt{-1} = i

so,

\frac{-5-\sqrt{48}i}{40}

Now solving:

=\frac{-5}{40}-\frac{i\,\sqrt{48}}{40}\\=\frac{-1}{8}-\frac{i\,\sqrt{2*2*2*2*3}}{40}\\=\frac{-1}{8}-\frac{i\,\sqrt{2^2*2^2*3}}{40}\\=\frac{-1}{8}-\frac{2*2\,i\,\sqrt{3}}{40}\\=\frac{-1}{8}-\frac{4\,i\,\sqrt{3}}{40}\\=\frac{-1}{8}-\frac{i\,\sqrt{3}}{10}

So, \frac{-5-\sqrt{-48}}{40}  in terms of i is: \frac{-1}{8}-\frac{i\,\sqrt{3}}{10}

3 0
3 years ago
Three boxes weighing 128 pounds each and one box weighing 254 pounds were loaded onto the back of an empty truck. A crate of app
Firlakuza [10]

Answer:

1352 pounds

Step-by-step explanation:

128+128=256

256+254=510

510+128=638

2000-638=1352

6 0
3 years ago
Nadia selects three marbles from the jar to the right without replacing them. What is the probability that all three marbles are
Savatey [412]
Not very likely. Does a picture come with this question.
4 0
3 years ago
3. What is the balance as a result of having a credit of $104 and a debit of $57?
Kruka [31]

Answer:

the answer is forty sevan

Step-by-step explanation:

104 - 57 equals 47

7 0
3 years ago
Other questions:
  • Julie is selling candy bars to raise money for new band uniforms. Candy bar X sells for $2 and candy bar Y sells for $3. Julie n
    13·2 answers
  • Solving two step equations
    15·2 answers
  • Joaquín invirtió su dinero a 12% y a 15% obteniendo unos intereses de $3000. Si las cantidades que invirtió hubieran sido interc
    5·1 answer
  • In the equation, 4x+14=-2x-22+12 , what are some of the steps that should be taken to solve for x? Select all that apply.
    6·1 answer
  • HELP PLEASE<br>Solve the compound inequality <br>3r + 2 &lt; 5 or 7r - 10 &gt; 60​
    12·1 answer
  • excavation scientists excavating a dinosaur mapped the site on a coordinate plane. if one bone lies from (-5,8) to (10,-1) and a
    7·1 answer
  • Genevieve bought a package of 16 sheets of parchment paper for $5.92.
    8·1 answer
  • Please help!! This is due today!!!
    8·1 answer
  • Jade used Fraction 3 over 4 yard of fabric to make a scarf. Can she make 2 of these scarves with Fraction 1 and 4 over 5 yards o
    15·2 answers
  • Brady has a bag with 6 black pens, 6 blue pens, and 2 red pens. If he grabs a pen at random, what is the probability that the pe
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!