# Prove by induction that n^3+5n is divisible by 3 for every natural number.

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Proof by induction has three core elements to it. To start with you must prove that the statement is true for the 'basic case'. For the most part this is 1, but some questions state it is higher.

Do this by subbing 1 into the equation and ensuring that it is divisible by 3

1^3 =1

5(1)=5

1+5=6 6/3=2 Therefore divisible by three and true for 1.

Then in order to futher prove it, we are going to assume that this is true for n=k

leaving us with the equation k^3+5k=3a as it is divisible by 3.

The next stage is to prove true for n=k+1.

Do this by subbing k+1 into the original equation:

(k+1)^3 +5(k+1)

multiplying this out gives:

k^3+3k^2+3k+1+5k+5

Now we have already established that k^3+5k=3a so through rearranging, k^3=3a-5k.

Subbing this into the k+1 equation gives us:

3d+3k^2+3k-6. Each element is a multiple of three so by taking three out leaves us:

3(d+k^2+k-2) which is a multiple of three and thus divisible by three.

Then leave a concluding statement along the lines of:

'As n^3+5n is true for n=k, then it is true for n=k+1. As it is true for n=1, then it must be true for n is greater than 1'

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