$$(2\frac{1}{3})^{2}-(\frac{1}{3})^{2}+1$$
$\frac{19}{3}\approx 6.333333333$
$$\left(\frac{6+1}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}+1$$
$$\left(\frac{7}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}+1$$
$$\frac{49}{9}-\left(\frac{1}{3}\right)^{2}+1$$
$$\frac{49}{9}-\frac{1}{9}+1$$
$$\frac{16}{3}+1$$
$$\frac{19}{3}$$
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$\frac{19}{3} = 6\frac{1}{3} = 6.333333333333333$
