# maths proofs

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Thread starter 16 years ago
#1
are these correct?
1.)
Prove that 2 consecutives numbers multiplied are even...
(2n) (2n+1) = 4n^2 + 2n

and since 2n + 2n proves that 2 evens added must be even and both the number in the above answer are even (both divide by 2) so will be even.

2.)
Prove that any 3 consecutive numbers are divisible by 3?
(x)(x+1)(x+2) = 3(x+1)

if the second one is correct then how do you get 3(x+1) from factorising the brackets?
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16 years ago
#2
(Original post by blag dahlia112)
are these correct?
1.)
Prove that 2 consecutives numbers multiplied are even...
(2n) (2n+1) = 4n^2 + 2n

and since 2n + 2n proves that 2 evens added must be even and both the number in the above answer are even (both divide by 2) so will be even.

2.)
Prove that any 3 consecutive numbers are divisible by 3?
(x)(x+1)(x+2) = 3(x+1)

if the second one is correct then how do you get 3(x+1) from factorising the brackets?
For the first one, just say that in the right hand side multiplying by 4 and 2 ensures an even number result.
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16 years ago
#3
(Original post by blag dahlia112)
are these correct?

2.)
Prove that any 3 consecutive numbers are divisible by 3?
(x)(x+1)(x+2) = 3(x+1)

if the second one is correct then how do you get 3(x+1) from factorising the brackets?
Provided 2) is written "Prove that the SUM of any 3 consecutive numbers is divisible by 3", then:

(x) + (x+1) + (x+2) = 3x + 3

= 3 (x+1)
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16 years ago
#4
In the second,

x(x+1)(x+2) = x(x^2+3x+3) = x^3+3x^2+3x and the multiplications by 3 ensure a result divisible by 3... I think...
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Thread starter 16 years ago
#5
what Expression sed is right so (hopefully) i just copied out the qu. wrong

ZJuwellH is wrong because of the addition which could make it wrong i think? because it would divide by 3 but then x^3 is added so i'm not sure if it would still divide by 3
0
16 years ago
#6
(Original post by blag dahlia112)
are these correct?
1.)
Prove that 2 consecutives numbers multiplied are even...
(2n) (2n+1) = 4n^2 + 2n

and since 2n + 2n proves that 2 evens added must be even and both the number in the above answer are even (both divide by 2) so will be even.

2.)
Prove that any 3 consecutive numbers are divisible by 3?
(x)(x+1)(x+2) = 3(x+1)

if the second one is correct then how do you get 3(x+1) from factorising the brackets?
Nice way to write your solution to the first one would simply be to take out the two giving 2(2n^2+n). Since this is divisible by two, it must be even.

If the second is the product of three consecutive numbers, then one of the three must be a multiple of three. Therefore, it will have 3 as a factor.
0
16 years ago
#7
(Original post by blag dahlia112)
what Expression sed is right so (hopefully) i just copied out the qu. wrong

ZJuwellH is wrong because of the addition which could make it wrong i think? because it would divide by 3 but then x^3 is added so i'm not sure if it would still divide by 3
You're right, I just went through it with x = 2 and it's wrong. theone's online now, better talk to him...
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Thread starter 16 years ago
#8
yeah ok i think i've got it all now

thanks

btw do you know any other possible proofs involving consecutive numbers which have appeared on past gcse papers?
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16 years ago
#9
(Original post by meepmeep)
Nice way to write your solution to the first one would simply be to take out the two giving 2(2n^2+n). Since this is divisible by two, it mus be even.

If the second is the product of three consecutive numbers, then one of the three must be a multiple of three. Therefore, it will have 3 as a factor.
Or the Fred Flinstone guy is pretty good too...
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Thread starter 16 years ago
#10
yeah i think expression got it right though and i probably wrote down the question one
0
16 years ago
#11
you could say that in any two consecutive numbers, one is even and one is odd, and an even times an odd is always even.
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