Prove: An odd number cubed is an odd number. n + (2n + 1)3 = [ 2 ]n3 + [ ]n2 + [ [](4n3 + 6n2 + 3n) + = an odd

Question

Prove: An odd number cubed is an
odd number.
n +
(2n + 1)3 = [ 2 ]n3 + [ ]n2 + [
[](4n3 + 6n2 + 3n) +
= an odd

in progress 0
Neala 5 years 2021-08-11T09:10:42+00:00 1 Answers 26 views 0

Answers ( )

  1. thongdat2
    0
    2021-08-11T09:12:14+00:00
    Reply

    Let n be any odd integer. Then n = 2k + 1 for some integer k.

    So

    n ³ = (2k + 1)³

    … = (2k)³ + 3 (2k)² + 3 (2k) + 1

    … = 8k ³ + 12k ² + 6k + 1

    … = 2 (4k ³ + 6k ² + 3k) + 1

    is also odd, since 4k ³ + 6k ² + 3k is just another integer.

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