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. 330
B. 165
C. 190
D. 225

Answer

A

Explanation

We have, `f(x)=ax^(2)+bx+candf(x+y)=f(x)+f(y)+xy`.
`:." "f(x+y)=f(x)+f(x)+f(y)+xy" for all "x,y inR`
`rArr" "a(x+y)^(2)+b(x+y)+cax^(2)+bx+c+ay^(2)+c+xy" for all "x,y in R`.
`rArr" "2axy=cxy" for all "x,yinR`
`rArr" "(2a-1)xy-c=0rArra=(1)/(2),c=0`
But a+b+c=3
`rArr" "(1)/(2)+b+0=3rArrb=(5)/(2)`
`:." "f(x)=ax^(2)+bx+c`
`rArr" "f(x)=(1)/(2)x^(2)+(5)/(2)x`
Hence, `underset(n=1)overset(10)sumf(n)=(1)/(2)underset(n=1)overset(10)sumn^(2)+(5)/(2)underset(n=1)overset(10)sumn=(1)/(2)xx(10(10+1)(20+1))/(6)+(5)/(2)xx(10(10+1))/(2)=330`