Using the direct proof, we assume that n>k=1n > k = 1.

n2>n>1n^2 > n > 1

f(n)=5n2+6n+7<5n2+6n2+7n2=18n2<20n2f( n ) = 5n^2+6n+7 < 5n^2+6n^2+7n^2 = 18n^2 < 20 n^2

f(n)=5n2+6n+7f( n ) = 5n^2+6n+7 is O(n2)O( n^2 )

To prove f(n)f( n ) is O(n2)O ( n^2 ), we use the constructive proof technique, and let k=1k=1 and C=20C=20.

proof

Using the UG rule, we consider an arbitrary integer nn.

We have shown that there exist kk and CC such that ∀n((n>k)→(f(n)<Cn2))\forall n (( n > k) \to (f( n ) < C n^2)), where n∈Zn \in \mathbb{Z}.

To prove ∀n((n>k)→(f(n)<Cn2))\forall n ((n > k) \to (f( n ) < C n^2)), where n∈Zn \in \mathbb{Z}

QED