JEE MAIN - Mathematics (2003 - No. 4)
If $$f:R \to R$$ satisfies $$f$$(x + y) = $$f$$(x) + $$f$$(y), for all x, y $$ \in $$ R and $$f$$(1) = 7, then $$\sum\limits_{r = 1}^n {f\left( r \right)} $$ is
$${{7n\left( {n + 1} \right)} \over 2}$$
$${{7n} \over 2}$$
$${{7\left( {n + 1} \right)} \over 2}$$
$$7n + \left( {n + 1} \right)$$
Explanation
$$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).$$
Function should be $$f(x)=mx$$
$$f\left( 1 \right) = 7;$$
$$\therefore$$ $$m=7,$$ $$f\left( x \right) = 7x$$
$$\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}} $$
Function should be $$f(x)=mx$$
$$f\left( 1 \right) = 7;$$
$$\therefore$$ $$m=7,$$ $$f\left( x \right) = 7x$$
$$\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}} $$
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