Proof by induction

  • #1
yankees26an
36
0

Homework Statement



Prove by induction, that for all integer n where n>= 1

[tex] \sum_{i=1}^{n} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4} [/tex]



Homework Equations





The Attempt at a Solution



First question is do I start at i=0 or i=1? It says >=, so not sure.

Ok then I added (n+1)(n+2)(n+3) to the right to balance out the n+1 on the left. But what do I do after that. Just simplify? Or am I missing something

[tex] \sum_{i=1}^{n+1} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3) [/tex]

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
VeeEight
615
0
Just add (n+1)(n+2)(n+3)=x to the fraction by multiplying it by 4/4, and factor x out of the resulting terms.
 
  • #3
yankees26an
36
0
Just add (n+1)(n+2)(n+3)=x to the fraction by multiplying it by 4/4, and factor x out of the resulting terms.

Ok, so basically just plugin n+1 for i on the left, and simplify the right. Set them equal, and done?
 
  • #4
VeeEight
615
0
You went to n+1 in your sum, took out the n+1 case and applied the induction hypothesis to the sum up to n. You are then left with [tex]
\sum_{i=1}^{n+1} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)
[/tex].

Now you need to take (n+1)(n+2)(n+3) = 4[(n+1)(n+2)(n+3] / 4 and add the two fractions together. Then, yes, you simplify and find it is equal to what you desired.
 
  • #5
yankees26an
36
0
You went to n+1 in your sum, took out the n+1 case and applied the induction hypothesis to the sum up to n. You are then left with [tex]
\sum_{i=1}^{n+1} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)
[/tex].

Now you need to take (n+1)(n+2)(n+3) = 4[(n+1)(n+2)(n+3] / 4 and add the two fractions together. Then, yes, you simplify and find it is equal to what you desired.

Did you take the (n+1)(n+2)(n+3) from the right side or from the i's? What happens to the extra (n+1)(n+2)(n+3) terms on the right?
 
  • #6
VeeEight
615
0
[tex]
\sum_{i=1}^{n+1} i(i+1)(i+2) =
(n+1)(n+2)(n+3) +
\sum_{i=1}^{n} i(i+1)(i+2)
[/tex]
[tex]= (n+1)(n+2)(n+3)+
\frac{n(n+1)(n+2)(n+3)}{4}
[/tex] (by the induction hypothesis)


[tex]
= \frac{4(n+1)(n+2)(n+3)}{4} +
\frac{n(n+1)(n+2)(n+3)}{4}

[/tex]

Add these fractions together and factor out (n+1)(n+2)(n+3)
 
  • #7
yankees26an
36
0
You completely lost me..

What's on the left side and the right side?
 
  • #8
yankees26an
36
0
Ok so I simplified and it's

1/4 (n+1) (n+2) (n+3) (n+4)

So what do I do after I simplify?
 
  • #9
VeeEight
615
0
Can you show me where I lost you?
You proved the base case. So that's done.
Now the goal is to show that for the statement S(n), S(n) is true implies S(n+1) is true.
So the goal in my last post was to take the sum to n+1 and try to reduce it to showing that it is equal to [(n+1)(n+2)(n+3)(n+4)]/4 (which is the n+1 case of the right hand right of the equality in your first post). In fact you had started this in your first post. You need to carry out all the simplification steps.
 
  • #10
VeeEight
615
0
Once you have achieved 1/4 (n+1) (n+2) (n+3) (n+4) after reducing from the original sum to n+1, you have finished the second step.
 

Suggested for: Proof by induction

  • Last Post
Replies
4
Views
161
  • Last Post
Replies
7
Views
438
  • Last Post
Replies
4
Views
566
Replies
6
Views
223
  • Last Post
Replies
2
Views
793
Replies
4
Views
144
  • Last Post
Replies
3
Views
913
  • Last Post
Replies
1
Views
899
  • Last Post
Replies
19
Views
195
Top