Question

Show that the sum of the cubes of
Consicutive integer divisible by 9​

Answer 1

Answer:

Let n∈Z, then we have to look at

(n−1)3+n3+(n+1)3=3n(n2+2)

so if we can show that n(n2+2) is divisible by 3 we are done. So I divide it up into three cases

n≡0 (mod 3) means we are done.

n≡1 (mod 3), then n2≡1 (mod 3) and then 3∣(n2+2) and we are done.

n≡2 (mod 3), then n2≡1 (mod 3) and then again 3∣(n2+2) and in this case too we get the desired result.

Answer 2

Answer:

Let p(n)=n³+(n+1)³+(n+2)³ is divisible by 9. hence

, by the principle of mathmatical induction the sum of cubes of three consecutive integers is divisible by 9i.e. p(n) is true for all n€N...