JEE Main Solved Paper 2017 Question 11
Question: 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,\forall x,y\in R, $ then $ \sum\limits _{n=1}^{10}{f(n)} $ is equal to: [JEE Main Solved Paper-2017]
Options:
A) 225
B) 330
C) 165
D) 190
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=ax^{2}+bx+c $ $ f(1)=a+b+c=3 $
Now $ f(x+y)=f(x)+f(y)+xy $
put $ y=1 $ $ f(x+1)=f(x)+f(1)+x $
$ f(x+1)=f(x)+x+3 $
Now $ f(2)=7 $ $ f(3)=12 $
Now $ S _{n}=3+7+12+……t _{n} $ ?..
$ S _{n}=3+7+……{t _{n-1}}+t _{n} $ ?..
On subtracting from $ t _{n}=3+4+5+….. $ upto n terms $ t _{n}=\frac{(n^{2}+5n)}{2} $ $ S _{n}=\sum{t _{n}}=\sum{\frac{(n^{2}+5n)}{2}} $
$ S _{n}=\frac{1}{2}[ \frac{n(n+1)(2n+1)}{6}+\frac{5n(n+1)}{2} ] $ $ S _{10}=330 $
