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 " underset (n=1)overset (10)(sum)f(n)` is equal to

Options

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

Answer

C

Explanation

We have , `f(x)=ax^(2)+bx+c`
`and , f(x+y)=f(x)+f(y)+xy`
`:. a(x+y)^(2)+b(x+y)+c=(ax^(2)+bx+c)+(ay^(2)+by+c)+xy` for all `x,y in R `
`implies (2a-1)xy-c=0` for all `x,y in R`
`implies 2a-1=0 and c=0`
`implies a=(1)/(2) and c=0`
Substituting these values in `a+b+c=3`, we get `b=(5)/(2)`
`:. f(x)=(1)/(2)x^(2)+(5)/(2)x`
Hence, `sum_(n=1)^(10) f(n)=sum_(n=1)^(10)((1)/(2)n^(2)+(5)/(2)n)=(1)/(2)sum_(n=1)^(10)n^(2)+(5)/(2)sum_(n=1)^(10)n`
`=(1)/(2)xx(10(10+1)(20+1))/(6)+(5)/(2)xx(10(10+1))/(2)=330`