$f\left( x \right) = a{x^2} + bx + c$

$f\left( 1 \right) = a + b + c = 3 \Rightarrow f\left( 1 \right) = 3$

Now $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy\,\,\,\,\,\,…\left( 1 \right)$

Put $x=y=1$ in eqn $(1)$

$f\left( 2 \right) = f\left( 1 \right) + f\left( 1 \right) + 1$

$\,\,\,\,\,\,\,\,\,\,\,\, = 2f\left( 1 \right) + 1$

$f\left( 2 \right) = 7$

$ \Rightarrow f\left( 3 \right) = 12$

Now, ${S_n} = 3 + 7 + 12 + ….{t_n}\,\,\,\,\,\,\,\,…..\left( 1 \right)$

${S_n} = 3 + 7 + ….{t_{n – 1}} + {t_n}\,\,\,\,\,\,\,……\,\left( 2 \right)$

Subtract $(2)$ from $(1)$

${t_n} = 3 + 4 + 5 + …upto\,n\,terms$

${t_n} = \frac{{\left( {{n^2} + 5n} \right)}}{2}$

${S_n} = \sum {{t_n} = \sum {\frac{{\left( {{n^2} + 5n} \right)}}{2}} } $

${S_n} = \frac{1}{2}\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{5n\left( {n + 1} \right)}}{2}} \right] = \frac{{n\left( {n + 1} \right)\left( {n + 8} \right)}}{6}$

${S_{10}} = \frac{{10 \times 11 \times 18}}{6} = 330$