Prove that n^3+5n is divisible by 6

To Prove

n³ + 5n is divisible by 6 ∀ n ∈ N

Proof

We will prove the given statement using the principle of mathematical induction

Step 1

We need to check if the statement is true for n = 1.

n³ + 5n = 1³ + 5(1) = 1 + 5 = 6

Therefore, the statement is true for n = 1.

Step 2

Let the statement be true for n = k

$k^3+5k$ is divisible by 6. We can write the statement as

$k^3+5k = 6x$ …(1)

Step 3

In this step, we need to prove that if the statement is true for n = k, then it is also true for n = k + 1.

$\\*n^3+ 5n = (k+1)^3 + 5(k+1)\\*n^3 + 5n = k^3 + 1 + 3k^2 + 3k + 5k + 5\\*n^3 + 5n = k^3 + 5k + 6 + 3k(k+1)\\*n^3 + 5n = (k^3 + 5k) + 6 + 3k(k+1)\\*n^3 + 5n = 6x + 6 + 3k(k+1) ...(2)$

Now, the term k(k+1) is always even. You can refer here for the proof.

So, we can write

$k(k+1) = 2y$

Substituting the above equation in (2), we get

$\\*n^3 + 5n = 6x + 6 + 6y\\*n^3 + 5n = 6(x + 1 + y)$

$6(x + 1 + y)$ is divisible by 6. Thus, the given statement is true for n = k + 1 if it is true for n = k. Therefore, by the principle of mathematic induction, the given statement “n³ + 5n is divisible by 6″ is true ∀ n ∈ N.

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