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 Sigma_(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
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)+x=f(x)+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+.... + l_(n-1)+l_n` (ii)
Subtracting (ii) from (i) , we get
`t_n=3+4+5+.....` upto n terms
`because t_n =((n+2)(n+3))/(2)-3`
`=(n^2 +5n)/(2)`
`S_n= Sigma t_n = Sigma (n^2+5n)/(2)`
`=1/2[(n(n+1)(2n+1))/(6)+(5n(n+1))/(2)]`
`=(n(n+1)(n+8))/(6)`
`therefore S_(10)=330`