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
JEE · Math · previous-year question
- A.$${{7n\left( {n + 1} \right)} \over 2}$$correct
- B.$${{7n} \over 2}$$
- C.$${{7\left( {n + 1} \right)} \over 2}$$
- D.$$7n + \left( {n + 1} \right)$$
Answer
A. $${{7n\left( {n + 1} \right)} \over 2}$$
Explanation
$$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).$$ Function should be $$f(x)=mx$$ $$f\left( 1 \right) = 7;$$ $$\therefore$$ $$m=7,$$ $$f\left( x \right) = 7x$$ $$\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}} $$
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