If \( \Large f:R \rightarrow R \) satisfies \( \Large f \left(x+y\right) = f \left(x\right) = f \left(x\right)+f \left(y\right) \) for all \( \Large x,\ y,\ \epsilon R\ and\ f \left(1\right)=7 \), then \( \Large \sum^{n}_{r=1}f \left(r\right) \) is:

A) \( \Large \frac{7n}{2} \)
B) \( \Large \frac{7 \left(n+1\right) }{2} \)

C) \( \Large 7n \left(n+1\right) \)
D) \( \Large \frac{7n \left(n+1\right) }{2} \)

Correct answer:
D) \( \Large \frac{7n \left(n+1\right) }{2} \)

Description for Correct answer:
\( \Large \sum_{r=1}^{n}f \left(r\right) = f \left(1\right)+f \left(2\right)+f \left(3\right)+....+f \left(n\right) \)

= \( \Large f\left(1\right)+2f \left(1\right)+3f \left(1\right)+...nf \left(1\right) \)

\( \Large \left[ since f \left(x+y\right)=f \left(x\right)+f \left(y\right) \right] \)

= \( \Large \left(1+2+3+...+n\right)f \left(1\right) \)

= \( \Large \frac{n \left(n+1\right) }{2}.7 = \frac{7n \left(n+1\right) }{2} \)

Please provide the error details in above question

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