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Question

Let $$a,b,c\varepsilon R$$. If $$f(x)=ax^2+bx+c$$ is such that $$a+b+c=3$$ and $$f(x+y)=f(x)+f(y), \forall\, x,y \varepsilon R$$, then $$\sum^{10}_{n=1}f(n)$$ is equal to


A
165
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B
190
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C
255
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D
336
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Solution

The correct option is D $$165$$
Given, $$f(x)=a{ x }^{ 2 }+bx+c$$,
            $$a+b+c=3.$$
Also, $$f(x+y)=f(x)+f(y)$$
    $$\Rightarrow a{ (x+y) }^{ 2 }+b(x+y)+c=(a{ x }^{ 2 }+bx+c)+(a{ y }^{ 2 }+by+c)$$
    $$\Rightarrow a({ x }^{ 2 }+{ y }^{ 2 }+2xy)+b(x+y)+c=a({ x }^{ 2 }+{ y }^{ 2 })+b(x+y)+2c$$
    $$\Rightarrow a({ x }^{ 2 }+{ y }^{ 2 })+2axy+b(x+y)+c=a({ x }^{ 2 }+{ y }^{ 2 })+b(x+y)+2c$$
    $$\Rightarrow 2axy-c=0$$
Now, Left-hand side gives $$0$$ for all real values of $$x$$ and $$y$$. Hence all coefficients must be $$0$$.
    $$\Rightarrow 2a=0\quad \& \quad c=0.$$
    $$\Rightarrow a=0\quad \& \quad c=0.$$
Since $$a+b+c=3$$,
     $$\Rightarrow 0+b+0=3$$
     $$\Rightarrow b=3$$

Therefore, $$f(x)=3x$$

Now, $$\sum _{ n=1 }^{ 10 }{ f(n) } $$
     $$=\sum _{ n=1 }^{ 10 }{ 3n } $$
     $$=3\sum _{ n=1 }^{ 10 }{ n } $$
     $$=3.\dfrac { 10(10+1) }{ 2 } $$                                          (Using $$\sum _{ r=1 }^{ n }{ r } =\dfrac { n(n+1) }{ 2 } $$)
     $$=165.$$

Hence, Option $$(A)$$ is correct.


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