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Proof by induction

  • #1

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
614
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
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
614
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
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
614
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
You completely lost me..

What's on the left side and the right side?
 
  • #8
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
614
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
614
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.
 

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