Using Mathematical Induction:
Prove the sum of cubes of 3 consecutive pos. integers is divisble by 9.
The sum of the cubes of 1, 2, 3 is 1+8+27=36 which is divisible by 9.
Now suppose that for some k: C(k)=k^3+(k+1)^3+(k+2)^3 is divisible by 9.
Then:
C(k+1)=C(k) - k^3 + (k+3)^3 = C(k) + 3 (k^2*3) + 3 (k*3^2) + 3^3
..............=C(k) + 9 [k^2 + 3k + 3]
hence as by assumption C(k) is divisible by 9 so is C(k+1), and as we have
established that C(1) is divisible by 9, we have established by mathematical
induction that C(n) is divisible by 9 for all positive integers n.
This proves (as it is the same thing in different words) that the sum of the
cubes of three consecutive positive integers is divisible by 9.
RonL
Hello, Ideasman!
This is Captain Black's proof . . . with my formatting.
Using Mathematical Induction:
Prove the sum of cubes of 3 consecutive pos. integers is divisble by 9.
Verify S(1): .1³ + 2³ + 3³ .= .36 . . . divisible by 9.
Assume S(k): .k³ + (k+1)³ + (k+2)³ .= .9a . for some integer a.
Add (k+3)³ - k³ to both sides:
. . k³ + (k+1)³ + (k+2)³ + (k+3)³ - k³ .= . 9a + (k+3)³ - k³
. . (k+1)³ + (k+2)³ + (k+3)³ .= .9a + k³ - 9k² + 27k - 27 - k³
. . (k+1)³ + (k+2)³ + (k+3)³ .= .9a - 9k² + 27k - 27
. . (k+1)³ + (k+2)³ + (k+3)³ .= .9(a - k² + 3k - 3)
The left side is the left side of S(k+1); the right side is a multiple of 9.
. . The inductive proof is complete.
Corrected my error . . . thanks, Captain!
Just a quick question-
I thought mathematicians don't like solving two solves of the equation at the same time to prove something... I thought you work with one side (IE: the (k + 1) side to show the other side; according to my math professor, it's bad to try solve a proof like this. I may be wrong..your thoughts?