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 $\displaystyle\sum^{10}_{n = 1} f(n)$ is equal to :

• A. 165
• B. 190
• C. 255
• D. 330

Answer 1

Answer: (D)

As, $f (x + y) = f (x) + f (y) + xy$
Given, $f (1) = 3$
Putting, $x = y = 1$
$\Rightarrow f (2) = 2f (1) +1 = 7$
Similarly, $x = 1, y = 2$
$\Rightarrow f (3) = f (1) + f (2) + 2 = 12$
Now, $\displaystyle \sum_{n=1}^{10}f\left(n\right) = f \left(1\right) + f \left(2\right) + f \left(3\right) + ... + f \left(10\right)$
$= 3 + 7 +12 + 18 + ... = S$ (let)
Now, $S_{n}$ = 3 + 7 + 12 + 18 + ... + $t_{n}$
Again, $S_{n}$ = 3 + 7 + 12 + ... + $t_{n-1} + t_{n}$
We ge, $t_{n}$ = 3 + 4 + 5+ ... n terms
$= \frac{n\left(n+5\right)}{2}$
i.e., $S_{n} = \displaystyle \sum_{n=1}^{n}t_{n} = \frac{1}{2}\left\{\sum n^{2}+5\sum n\right\}$
$ = \frac{n\left(n+1\right)\left(n+8\right)}{6}$
So, $S_{10} = \frac{10\times11\times18}{6} = 330$