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Question: Let we have \(a,b,c \in R\) if \(f(x) = a{x^2} + bx + c\) is such that \(a + b + c = 3\) and \(f(x +...

Let we have a,b,c∈Ra,b,c \in R if f(x)=ax2+bx+cf(x) = a{x^2} + bx + c is such that a+b+c=3a + b + c = 3 and f(x+y)=f(x)+f(y)+xy∀y∈Rf(x + y) = f(x) + f(y) + xy\forall y \in R. Then ∑n=110f(n)\sum\limits_{n = 1}^{10} {f(n)} is equal to.
A. 330
B. 165
C. 190
D. 255

Explanation

Solution

In order to solve this problem we need to know that we need to find the value of a, b, c by using differentiation and partial differentiation. We first need to find the value of a and b with the help of the information given to get the exact function f(x). Doing this and proceeding will give you the right answer.

Complete step-by-step solution:
It is given that,
f(x)=ax2+bx+cf(x) = a{x^2} + bx + c
And f(x+y)=f(x)+f(y)+xy∀y∈Rf(x + y) = f(x) + f(y) + xy\forall y \in R
And a+b+c=3a + b + c = 3
On differentiating f(x) with respect to x we get,
f(x)=2ax+bf(x) = 2ax + b
f’(0)=bf’(0) = b
since f(x+y)=f(x)+f(y)+xyf(x+y) = f(x) + f(y) + xy
Then we put x = y = 0
f(0+0)=f(0)+f(0)+0f(0+0) = f(0) + f(0) + 0
That is f(0)=0f(0) = 0
Partial differentiation w.r.t x we get,
f’(x+y)=f’(x)+yf’(x+y) = f’(x) + y
Put x = 0 in this equation.
Then we get,
f‘(y)=f’(0)+yf‘(y) = f’(0) + y
f’(y)=b’+y(f’(0)=b)f’(y) = b’ + y (f’(0) = b)
On integrating we get,
f(y)=b’y+y22+c′f(y) = b’y + \dfrac{{{y^2}}}{2} + c'
On putting y = x we get
f(x)=x22+bx+c’f(x) = \dfrac{{{x^2}}}{2} + bx + c’
Put x = 0 we get,
f(0)=c’=0f(0) = c’ = 0
On comparing with given function the given function with f(x) we get,
a=12,b’=b,c’=c=0a = \dfrac{1}{2}, b’ = b, c’ = c = 0.
Then the equation a+b+c=3a + b + c = 3 can be 12+b’+0=3\dfrac{1}{2} + b’ + 0 = 3
Then the value of b’ = 52\dfrac{5}{2}.
Then the equation f(x) we get as,
f(x)=12x2+52xf(x) = \dfrac{1}{2}{x^2} + \dfrac{5}{2}x
Now we have to find ∑n=110f(n)\sum\limits_{n = 1}^{10} {f(n)}
∑n=110f(n)=∑12n2+∑52n\sum\limits_{n = 1}^{10} {f(n)} = \sum {\dfrac{1}{2}{n^2} + \sum {\dfrac{5}{2}n} }
So on opening the summation we get the value as,
∑n=110f(n)=12n(n+1)(2n+1)6+52n(n+1)2\sum\limits_{n = 1}^{10} {f(n)} = \dfrac{1}{2}\dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{5}{2}\dfrac{{n(n + 1)}}{2}
On putting n= 10 we will get the value of ∑n=110f(n)\sum\limits_{n = 1}^{10} {f(n)}
∑n=110f(n)=1210(10+1)(2(10)+1)6+5210(10+1)2\sum\limits_{n = 1}^{10} {f(n)} = \dfrac{1}{2}\dfrac{{10(10 + 1)(2(10) + 1)}}{6} + \dfrac{5}{2}\dfrac{{10(10 + 1)}}{2}
On solving it we get,
∑n=110f(n)=192.5+137.5\sum\limits_{n = 1}^{10} {f(n)} = 192.5 + 137.5
∑n=110f(n)=330\sum\limits_{n = 1}^{10} {f(n)} = 330 is the answer to this problem.
Hence option A is the correct answer.

Note: Whenever we face such a type of question and always try to solve with partial differentiation it becomes very easy to tackle the question. First, always try to find the equation then apply the property of summation. You need to know that the sum of n natural number is n(n+1)2\dfrac{{n(n + 1)}}{2} and that of n2{n^2} is n(n+1)(2n+1)6\dfrac{{n(n + 1)(2n + 1)}}{6}. Knowing this will help you to solve the problems further.