SOLUTION: Prove that n^3 + (n+1)^3 + (n+2)^3 is divisible by 9 for all n in Natural numbers. I have tried, let mathematical induction, let P(n) be the statement, then then n^9 - n is divisi

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Question 97923This question is from textbook Intro to Real Analysis
: Prove that n^3 + (n+1)^3 + (n+2)^3 is divisible by 9 for all n in Natural numbers. I have tried, let mathematical induction, let P(n) be the statement, then then n^9 - n is divisible by 9. I'm stuck...
Thanks This question is from textbook Intro to Real Analysis

Found 2 solutions by aaaaaaaa, mathslover:
Answer by aaaaaaaa(138) About Me  (Show Source):
You can put this solution on YOUR website!
"x mod y" is short for "remainder of the division of x by y". Iff "x mod y" is 0, x is divisible by y. In our case, y = 9.

n%5E3+%2B+%28n%2B1%29%5E3+%2B+%28n%2B2%29%5E3 (mod 9) =
n%5E3+%2B+n%5E3+%2B+3n%5E2+%2B+3n+%2B+1+%2B+n%5E3+%2B+6n%5E2+%2B+12n+%2B+8 (mod 9)=
3n%5E3+%2B+9n%5E2+%2B+15n+%2B+9 (mod 9) =
3n%5E3+%2B+9n%5E2+%2B+9n+%2B+6n+%2B+9 (mod 9)
9n^2 is always divisible by 9, so we shouldn't worry about that term. The same goes for the "+9" at the end, and the "+9n" at the middle.
3n%5E3+%2B+cross%289n%5E2%29+%2B+cross%289n%29+%2B+6n+%2B+cross%289%29 (mod 9) =
3n%5E3+%2B+6n (mod 9)

Therefore,
P(x) = n%5E3+%2B+%28n%2B1%29%5E3+%2B+%28n%2B2%29%5E3 (mod 9) = 0
is equivalent to
S(x) = 3n%5E3+%2B+6n (mod 9) = 0

Now suppose S(n) is true. I'll try to prove that S(n+1) is also true.
3%28n%2B1%29%5E3+%2B+6%28n%2B1%29 (mod 9) =
3%28n%5E3+%2B+3n%5E2+%2B+3n+%2B+1%29+%2B+6n+%2B+15 (mod 9) =
3n%5E3+%2B+cross%289n%5E2%29+%2B+cross%289n%29+%2B+3+%2B+6n+%2B+15 (mod 9) =
3n%5E3+%2B+6n+%2B+cross%2818%29 (mod 9) =
3n%5E3+%2B+6n (mod 9)
But wait, if S(n) is true, then
3n%5E3+%2B+6n (mod 9) = 0 (mod 9)

The inductive step is done. Now we need to check the base case of natural numbers (which is 0)
3%280%5E3%29+%2B+3%2A0 (mod 9) =
0 (mod 9)
S(x) is therefore true for all naturals.

Answer by mathslover(157) About Me  (Show Source):
You can put this solution on YOUR website!
We prove it through mathematical induction
for n =1
P(n)= 1^3 + 2^3 + 3^3
= 36 which is divisble by 9
hence P(1) is true
Let it be true for n=m
P(m) is true
We have to now prove that it is true for n=m+1


given P(m+1) = (m+1)^3 + (m+2)^3 + (m+3)^3
= (m+1)^3 + (m+2)^3 + m^3 + 9m^2 + 27m + 27 Expanding the last term
= [ m^3 + (m+1)^3 + (m+2)^3 ] + 9m^2 + 27m + 27 Rearranging and grouping
= P(m) + 9(m^2+ 3m +3)
the expression above is also divisible by 9 as P(m) is divisble by 9 and
9(m^2+ 3m +3) is divisble by 9.
Hence it is true for all n . Proof by induction complete