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

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(B) Given $f(x) = ax^2 + bx + c$ and $a + b + c = 3$.Since $f(1) = a + b + c$,we have $f(1) = 3$.Given $f(x + y) = f(x) + f(y) + xy$.Putting $x=0, y=0

Vedclass · Accepted Answer

$330$

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(B) Given $f(x) = ax^2 + bx + c$ and $a + b + c = 3$.Since $f(1) = a + b + c$,we have $f(1) = 3$.Given $f(x + y) = f(x) + f(y) + xy$.Putting $x=0, y=0$,we get $f(0) = f(0) + f(0) + 0 \Rightarrow f(0) = 0$. Thus $c = 0$.Now,$f(x) = ax^2 + bx$.$f(1) = a + b = 3$.Using $f(x+y) = a(x+y)^2 + b(x+y) = ax^2 + ay^2 + 2axy + bx + by$.Also $f(x) + f(y) + xy = ax^2 + bx + ay^2 + by + xy$.Comparing coefficients of $xy$,we get $2a = 1 \Rightarrow a = 1/2$.Since $a + b = 3$,$b = 3 - 1/2 = 5/2$.So,$f(n) = \frac{1}{2}n^2 + \frac{5}{2}n = \frac{n^2 + 5n}{2}$.We need to find $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} \frac{n^2 + 5n}{2} = \frac{1}{2} [\sum_{n=1}^{10} n^2 + 5 \sum_{n=1}^{10} n]$.Using sum formulas: $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$.For $n=10$: $\sum n^2 = \frac{10 	imes 11 	imes 21}{6} = 385$ and $\sum n = \frac{10 	imes 11}{2} = 55$.Sum $= \frac{1}{2} [385 + 5(55)] = \frac{1}{2} [385 + 275] = \frac{660}{2} = 330$.

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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