Question

Use mathematical induction to prove the statement is true for all positive integers n.

The integer n3 + 2n is divisible by 3 for every positive integer n.

Answer 1

have you ever done a proof by induction?

Answer 2

Note that:
\(\color{#000000 }{ \displaystyle n^3+2n=n(n^2+2) }\)

we will split this into 3 categories:
\(\color{#000000 }{ \displaystyle n\in 3k }\)
\(\color{#000000 }{ \displaystyle n\in 3k+1 }\)
\(\color{#000000 }{ \displaystyle n\in 3k+2 }\)

Case 1:
\(\color{#000000 }{ \displaystyle n^3+2n=n(n^2+2) \quad \Longrightarrow \quad (3k)\cdot [(3k)^2+2)] }\)
(product of integers multiplied times 3, so that should be obvious)

Case 2:
\(\color{#000000 }{ \displaystyle n^3+2n \quad \Longrightarrow \quad (3k+1)^3+2(3k+1)}\)
(expand and keep in mind that k is a natural number)

Case 3:
\(\color{#000000 }{ \displaystyle n^3+2n \quad \Longrightarrow \quad (3k+2)^3+2(3k+2)}\)
(expand and keep in mind that k is a natural number)

For cases 2 and 3, you are just expanding, and you will get an expression that will be divisible by 3 for any positive integer k.