Natasha1
- 494
- 9
Which cube members are not in the sequence and prove it?
2, 5, 8, 11, 14, ...
How can this be proved
My answer:
an = 3n + 2
Any natural number may be written as N= 3k+p for some natural number K and p=0,1 or 2.
So
N^3=(3k+p)3
N^3=3(9k+k^2p+kp^2)+p^3
N^3=3k+p^3
Therefore N^3 (mod 3)
So as an is congruent to 2(mod 3), N^3 lies in the sequence. In other words, the cude of any number in the sequence is in the sequence. But how do I go and show which cube members are not in the sequence and moreover how do I prove it?
2, 5, 8, 11, 14, ...
How can this be proved

My answer:
an = 3n + 2
Any natural number may be written as N= 3k+p for some natural number K and p=0,1 or 2.
So
N^3=(3k+p)3
N^3=3(9k+k^2p+kp^2)+p^3
N^3=3k+p^3
Therefore N^3 (mod 3)
So as an is congruent to 2(mod 3), N^3 lies in the sequence. In other words, the cude of any number in the sequence is in the sequence. But how do I go and show which cube members are not in the sequence and moreover how do I prove it?