Let `a,b, c in R. " If " f(x)=ax^(2)+bx+c` is such that `a+b+c=3 and f(x+y)=f(x)+f(y)+xy, AA x,y in R, " then " sum_(n=1)^(10) f(n)` is equal to

Options

A. 255
B. 330
C. 165
D. 190

Answer

B

Explanation

`f(x)=ax^(2)+bx+c`
`f(1)=a+b+c=3` (given)
Now ` f(x+y)=f(x)+f(y)+xy`
Putting `y=1`, we get
`f(x+1)=f(x)+f(1)+x=f(x)+x+3`
Putting `x=1,` we get `f(2)=f(1)+1+3=7`
Putting `x=2,` we get `f(3)=f(2)+2+3=12` and so on.
Now `S_(10)=3+7+12+ ... +t_(n) " ...(i)" `
`S_(10)=3+7+ ...t_(n-1)+t_(n) " ...(ii)" `
Subtracting (ii) and (i), we get
`t_(n)=3+4+5+ ...` upto n terms
` :. t_(n)=((n+2)(n+3))/(2)-3=(n^(2)+5n)/(2)`
`S_(n)=sumt_(n)=sum(n^(2)+5n)/(2)`
`=(1)/(2)[(n(n+1)(2n+1))/(6)+(5n(n+1))/(2)]`
`=(n(n+1)(n+8))/(6)`
` :. S_(10)=330`