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(B) Given the functional equation $f(x+y) = f(x) + f(y) + xy$. Let $f(x) = ax^2 + bx + c$. Substituting this into the equation: $a(x+y)^2 + b(x+y) + c

Vedclass · Accepted Answer

$275$

Vedclass · Answer

(B) Given the functional equation $f(x+y) = f(x) + f(y) + xy$. Let $f(x) = ax^2 + bx + c$. Substituting this into the equation: $a(x+y)^2 + b(x+y) + c = (ax^2 + bx + c) + (ay^2 + by + c) + xy$. $a(x^2 + 2xy + y^2) + bx + by + c = ax^2 + ay^2 + bx + by + 2c + xy$. $ax^2 + 2axy + ay^2 + bx + by + c = ax^2 + ay^2 + bx + by + 2c + xy$. Comparing coefficients of $xy$,we get $2a = 1$,so $a = 1/2$. Comparing constant terms,$c = 2c$,so $c = 0$. Given $f(1) = 2$,we have $a(1)^2 + b(1) = 2$,so $1/2 + b = 2$,which gives $b = 3/2$. Thus,$f(x) = \frac{1}{2}x^2 + \frac{3}{2}x = \frac{x(x+3)}{2}$. We need to calculate $\sum_{k=1}^{10} f(k) = \sum_{k=1}^{10} (\frac{1}{2}k^2 + \frac{3}{2}k) = \frac{1}{2} \sum_{k=1}^{10} k^2 + \frac{3}{2} \sum_{k=1}^{10} k$. Using summation formulas: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. For $n=10$: $\sum k^2 = \frac{10(11)(21)}{6} = 385$ and $\sum k = \frac{10(11)}{2} = 55$. Sum $= \frac{1}{2}(385) + \frac{3}{2}(55) = 192.5 + 82.5 = 275$.

Let $f: N \rightarrow N$ be a function such that $f(x+y)=f(x)+f(y)+xy$ for every $x, y \in N$. If $f(1)=2$,then $\sum_{k=1}^{10} f(k)=$

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