Can mathematical induction prove this proposition?

In summary, the proposition states that the sum of the series 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2) is equal to [n(n+1)(n+2)(n+3)]/4. This has been proven by showing that if n=1, the left hand side (LHS) and right hand side (RHS) are equal, and if Pk is true, then P(k+1) is also true. This is shown by using the common denominator of 4 and factoring out the terms to show that the LHS and RHS are equal. Therefore, the proposition is proven to be true.
  • #1
L²Cc
149
0
Proposition: 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2) = [n(n+1)(n+2)(n+3)]/4

Step (1): If n=1 then LHS (left hand side) = 6, and RHS = 6
Thus, P1 is true.

Step (2): If Pk is true then
k(k+1)(k+2) = [k(k+1)(k+2)(k+3)]/4

Now,
k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

k(k+1)(k+2) +[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + (k+1)(k+2)(k+3)

[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4 ---> Common denominator - 4.

Then...what do i do? any clues? I have not proven my proposition, yet.
 
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  • #2
L²Cc said:
...

Now,
[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

Think about this equality. Does it make sense?
 
  • #3
oopsies i meant to write
k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)
 
  • #4
L²Cc said:
oopsies i meant to write
k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

Right, and that implies [tex]k(k+1)(k+2) + (k+1)(k+2)(k+3) = \frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)}{4}[/tex]. The rest is pretty obvious.
 
  • #5
im not very good at factoring, but ill give it a try and you see whether I am on the right track:
let k+1= A, k+2= B, k+3= C
k(A)(B)(C) + 4(A)(B)(C)/4
A(K+4)B(K+4)C(K+4)
Then,
(K+4)(ABC)
am i on the right track?
Then,
(k+4)(k+1)(k+2)(k+3)/4
Then? How do i prove that P(k+1) is true whenever Pk and P1 are true?!
 
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1. What is mathematical induction?

Mathematical induction is a proof technique used to establish the truth of a statement for all natural numbers. It relies on the principle that if the statement is true for one number, and it can be shown that it must also be true for the next consecutive number, then it must be true for all numbers.

2. How does mathematical induction work?

Mathematical induction works by first proving a base case, which is typically the statement for the number 1. Then, using the induction hypothesis, it is shown that if the statement holds true for some number n, it must also be true for the next consecutive number n+1. This process is repeated until it can be concluded that the statement is true for all natural numbers.

3. What is the difference between strong and weak induction?

In strong induction, the induction hypothesis is that the statement is true for all numbers less than or equal to n. This allows for a stronger base case, as it can be assumed that the statement is true for all numbers leading up to n. In weak induction, the induction hypothesis is that the statement is true for n, and the proof must rely on the truth of the statement for n-1.

4. What are some common examples of mathematical induction?

Some common examples of mathematical induction include proving that a statement is true for all natural numbers, such as the sum of the first n natural numbers being n(n+1)/2. It can also be used to prove properties of sequences and series, and to solve problems in combinatorics and set theory.

5. When should mathematical induction be used?

Mathematical induction should be used when the statement being proven involves natural numbers or integers, and can be broken down into smaller, consecutive steps. It is also useful in proving properties of recursive algorithms and data structures. However, it should not be used as a substitute for direct proof, as some statements may be easier to prove using other techniques.

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