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Question

Let a, b, cR. If f(x)=ax2+bx+c is such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy, x, yR, then 10n=1f(n) is equal to:

A
330
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B
165
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C
190
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D
255
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Solution

The correct option is A 330
f(x)=ax2+bx+c
and f(x+y)=f(x)+f(y)+xy
a(x+y)2+b(x+y)+c=(ax2+bx+c)+(ay2+by+c)+xy
2axy=xy+c
a=12 and c=0
Also, a+b+c=3b=52
f(x)=12x2+52x

Then 10n=1f(n)=1210n=1n2+5210n=1n
=12×10×11×216+52×10×112
=330

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