Let a,b,c∈R. If f(x)=ax2+bx+c is such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy,∀x,y∈R, then 10∑n=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 A330 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=3⇒b=52 ∴f(x)=12x2+52x
Then 10∑n=1f(n)=1210∑n=1n2+5210∑n=1n =12×10×11×216+52×10×112 =330