Mathematica Mathematical Induction

  • Thread starter L²Cc
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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.
 
Last edited:

radou

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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?
 
149
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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)
 

radou

Homework Helper
3,105
6
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.
 
149
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im not very good at factoring, but ill give it a try and you see whether im 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???!
 
Last edited:

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