Let a,b,cϵR. If f(x)=ax2+bx+c be such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy,∀x,yϵR, then ∑10n=1f(n) is equal to
A
330
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
165
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
190
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
255
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 330 We have, f(x)=ax2+bx+c Now,f(x+y)=f(x)+f(y)+xy Puty=0⇒f(x)=f(x)+f(0)+0 ⇒f(0)=0 ⇒c=0 Again, puty=−x ∴f(0)=f(x)+f(−x)−x2 ⇒0=ax2+bx+ax2−bx−x2⇒2ax2−x2=0⇒a=12 Also,a+b+c=3 ⇒12+b+0=3⇒b=52∴f(x)=x2+5x2Now,f(n)=n2+5n2=12n2+52n ∴∑10n=1f(n)=12∑10n=1n2+52∑10n=1n=12⋅10×11×216+52⋅10×112=3852+2752=6602=330