Can anyone help me confirm if I have solved this correctly?

Many thanks.3

Q.^{n}> n^{2}for n>2, n $\displaystyle \in \mathbb{N}$.

Attempt:Step 1:For n = 2,

3^{2}= 9 & 2^{2 }= 4

Since 9 > 4, the statement is true for n = 2.

Step 2:Assume the statement is true for n = k, i.e. assume 3^{k}> k^{2}.

We must now show that the statement is true for n = k + 1,

i.e. 3^{k+1 }> (k + 1)^{2}

3^{k+1}= 3^{1}$\displaystyle \cdot$3^{k}...(3^{k}> k^{2}...assumed)

If it can be shown that 3(k^{2}) > (k + 1)^{2}, then 3^{k+1}> (k + 1)^{2}

3(k^{2}) > (k + 1)^{2 }if 3k^{2}- (k + 1)^{2}> 0

if 3k^{2}- k^{2}- 2k - 1 > 0

if 2k^{2}- 2k - 1 > 0

if 2k^{2}- 2k + 1 - 1 - 1 > 0...completing the square

if 2k^{2}- 2k + 1 - 2 > 0

if (2k - 1)(k - 1) - 2 > 0...true for k>2 = 3^{k+1}> (k + 1)^{2}

Therefore, the statement is true for n = k + 1, if true for n = k. Thus the statement is true for all n>2, n $\displaystyle \in \mathbb{N}$.