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Let $a,b,c\; \in R.$ If $f\left( x \right) = a{x^2} + bx + c$ is such that $a + b + c = 3$ and $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy,$ $\forall x,y \in R,$ then $\mathop \sum \limits_{n = 1}^{10} f\left( n \right)$ is equal to :
$255$
$330$
$165$
$190$
Solution
$f\left( x \right) = a{x^2} + bx + c$
$f\left( 1 \right) = a + b + c = 3 \Rightarrow f\left( 1 \right) = 3$
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}} } $
${S_n} = \frac{1}{2}\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{5n\left( {n + 1} \right)}}{2}} \right] = \frac{{n\left( {n + 1} \right)\left( {n + 8} \right)}}{6}$
${S_{10}} = \frac{{10 \times 11 \times 18}}{6} = 330$