Let a, b, c in R. If f(x) = ax^2 + bx + c is | vedclass

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(B) Given $f(x) = ax^2 + bx + c$. Since $f(x+y) = f(x) + f(y) + xy$,putting $x=0, y=0$ gives $f(0) = f(0) + f(0) + 0$,so $f(0) = 0$. Thus,$c = 0$. Giv

Vedclass · Accepted Answer

$330$

Vedclass · Answer

(B) Given $f(x) = ax^2 + bx + c$. Since $f(x+y) = f(x) + f(y) + xy$,putting $x=0, y=0$ gives $f(0) = f(0) + f(0) + 0$,so $f(0) = 0$. Thus,$c = 0$. Given $a+b+c = 3$,and since $c=0$,we have $a+b = 3$. Also,$f(x+y) = a(x+y)^2 + b(x+y) = a(x^2 + 2xy + y^2) + bx + by = (ax^2 + bx) + (ay^2 + by) + 2axy$. Comparing this with $f(x+y) = f(x) + f(y) + xy$,we get $2axy = xy$,which implies $2a = 1$,so $a = 1/2$. Since $a+b = 3$,$b = 3 - 1/2 = 5/2$. Thus,$f(n) = \frac{1}{2}n^2 + \frac{5}{2}n = \frac{n^2 + 5n}{2}$. Now,$\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} \frac{n^2 + 5n}{2} = \frac{1}{2} \left[ \sum_{n=1}^{10} n^2 + 5 \sum_{n=1}^{10} n ight]$. Using the formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$: $\sum_{n=1}^{10} n^2 = \frac{10 	imes 11 	imes 21}{6} = 385$. $\sum_{n=1}^{10} n = \frac{10 	imes 11}{2} = 55$. Therefore,$\sum_{n=1}^{10} f(n) = \frac{1}{2} [385 + 5(55)] = \frac{1}{2} [385 + 275] = \frac{660}{2} = 330$.

Let $a, b, c \in R$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy$ for all $x, y \in R$,then $\sum_{n=1}^{10} f(n)$ is equal to:

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