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Question

Let a,b,cϵR. If f(x)=ax2+bx+c be such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy, x, yϵR, then 10n=1 f(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
We have, f(x)=ax2+bx+c
Now, f(x+y)=f(x)+f(y)+xy
Put y=0f(x)=f(x)+f(0)+0
f(0)=0
c=0

Again, put y=x
f(0)=f(x)+f(x)x2
0=ax2+bx+ax2bxx2 2ax2x2=0 a=12
Also, a+b+c=3
12+b+0=3b=52 f(x)=x2+5x2Now, f(n)=n2+5n2=12n2+52n
10n=1 f(n)=12 10n=1 n2+5210n=1 n=1210×11×216+5210×112=3852+2752=6602=330

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