SOLUTION: prove that the product of any three consecutive numbers is even.
So far I have tried , (n)(n+1)(n+2)=n^3+3n^2+2n
but there is no multiple of an even number. I'm not sure wh
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-> SOLUTION: prove that the product of any three consecutive numbers is even.
So far I have tried , (n)(n+1)(n+2)=n^3+3n^2+2n
but there is no multiple of an even number. I'm not sure wh
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Question 965979: prove that the product of any three consecutive numbers is even.
So far I have tried , (n)(n+1)(n+2)=n^3+3n^2+2n
but there is no multiple of an even number. I'm not sure what kind of proof method it is but this would be a proof with the same method. i.e prove that the sum of three consecutive odd numbers is divisible by 3 , 2n+1+2n+3+2n+5= 6n+9 3(2n+3) is a multiple of three therefore it is divisible by three.
Any help would be appreciated, hope it is not too confusing
Thanks
You can put this solution on YOUR website! You should be allowed to use simple logic of two cases. Either ONE of the numbers is EVEN or TWO of the numbers are EVEN. In either case, the result will be even. Any product of integers which contains a multiple of two will be also a multiple of two; meaning that the product is even.
You can put this solution on YOUR website! let the consecutive numbers are n,(n+1),(n+2)
product = n*(n+1)*(n+2)
if n=1
P(1) = 1*2*3=6 --Even
Assume that P(n)=even then we have to prove that P(n+1)=(n+1)*(n+2)*(n+3) is also even
P(n+1)= (n+1)*(n+2)*(n+3)
=n*(n+1)*(n+2)+ 3*(n+1)*(n+2)
we know that n*(n+1)*(n+2)is even ( assumed )
we have to show that 3*(n+1)*(n+2) is also even
if n is odd then n+1 is even
either way n^2+3*n+2 is even
so 3*(n+1)*(n+2)=3*(n^2+3*n+2) is also even
we have shown that P(1) is even and that P(n+1)is even if P(n)is even.
hence proved.
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