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Use the Principle of Mathematical Induction to show that the following statement is true for all natural numbers n.
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\[P(n)=n ^{3}+2n+6\]
What is the statement..?
\[P(n)=n ^{3}+2n+6\]
is divisible by 3.
ah, there we go. Okay, so lets say P(n) is divisible by three. We seek to show that this implies that P(n+1) is divisible by three.
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\[P(n+1) = (n+1)^3 + 2(n+1) + 6 = n^3 + 3n^2 + 3n + 1 + 2n + 2 + 6\] \[= (n^3+2n+6) + 3n^2 + 3n +3 = P(n) + 3(n^2+n+1)\] So surely if P(n) is divisible by three, P(n+1) is divisible by three, right? Now all that remains is to prove it for n = 1. P(1) = 1 + 2+ 6 = 9 which is divisible by three. So, by induction, it is true for any natural number n.
Thank you.
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