ધારો કે a,b,c; in R. જો fleft( x right) = a{x | vedclass

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Question

$f\left( x \right) = a{x^2} + bx + c$

$f\left( 1 \right) = a + b + c = 3 \Rightarrow f\left( 1 \right) = 3$

Now $f\left( {x + y} \right) =

Vedclass · Accepted Answer

$330$

Vedclass · Answer

$f\left( x ight) = a{x^2} + bx + c$

$f\left( 1 ight) = a + b + c = 3 \Rightarrow f\left( 1 ight) = 3$

Now $f\left( {x + y} ight) = f\left( x ight) + f\left( y ight) + xy\,\,\,\,\,\,...\left( 1 ight)$

Put $x=y=1$ in eqn $(1)$

$f\left( 2 ight) = f\left( 1 ight) + f\left( 1 ight) + 1$

$\,\,\,\,\,\,\,\,\,\,\,\, = 2f\left( 1 ight) + 1$

$f\left( 2 ight) = 7$

$ \Rightarrow f\left( 3 ight) = 12$

Now, ${S_n} = 3 + 7 + 12 + ....{t_n}\,\,\,\,\,\,\,\,.....\left( 1 ight)$

${S_n} = 3 + 7 + ....{t_{n - 1}} + {t_n}\,\,\,\,\,\,\,......\,\left( 2 ight)$

Subtract $(2)$ from $(1)$

${t_n} = 3 + 4 + 5 + ...upto\,n\,terms$

${t_n} = \frac{{\left( {{n^2} + 5n} ight)}}{2}$

${S_n} = \sum {{t_n} = \sum {\frac{{\left( {{n^2} + 5n} ight)}}{2}} } $

${S_n} = \frac{1}{2}\left[ {\frac{{n\left( {n + 1} ight)\left( {2n + 1} ight)}}{6} + \frac{{5n\left( {n + 1} ight)}}{2}} ight] = \frac{{n\left( {n + 1} ight)\left( {n + 8} ight)}}{6}$

${S_{10}} = \frac{{10 	imes 11 	imes 18}}{6} = 330$

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