Question
by taking three different values on n, show that a cube of an even natural number n, is always even end cube of an odd natural number n, is always odd.
Answers
Steve
n even:
n = 2k
n^3 = (2k)^3 = 8k^3
an even multiple of a number is always even
n odd:
n = (2k+1)
n^3 = 8k^3 + 4k^2 + 2k + 1
= 2(4k^3 + 2k^2 + k) + 1
The extra +1 at the end makes the result odd.
n = 2k
n^3 = (2k)^3 = 8k^3
an even multiple of a number is always even
n odd:
n = (2k+1)
n^3 = 8k^3 + 4k^2 + 2k + 1
= 2(4k^3 + 2k^2 + k) + 1
The extra +1 at the end makes the result odd.