If $f:[R \rightarrow[R$ is such that $f(x+y)=f(x)+f(y)$ for all $x, y \in[R . f(\mathrm{l})=7$ and $\sum_{r=1}^n f(r)=14112$, then $n=$

Question

Aicharya · Accepted Answer

We have, $f(x+y)=f(x)+f(y)$ and $f(1)=7$ $ \begin{array}{rlrl} \because & & f(1) & =7 \ & 	herefore & & f(2)=f(1)+f(1) \ & & f(2)=7+7=14 \end{array} $ Similarly, $ \begin{aligned} & f(x)=f(1)+f(2)=7+14=21 \ & f(4)=f(1)+f(3)=7+21=28 \ & f(n)=7 n \end{aligned} $ $ 	herefore \quad f(n)=7 n $ Now, $\quad \sum_{r=1}^n f(r)=14112$ $ \begin{array}{lr} \Rightarrow & f(1)+f(2)+f(3)+\ldots+f(n)=14112 \ \Rightarrow & 7+14+21+\ldots+7 n=14112 \ \Rightarrow & 7(1+2+3+\ldots+n)=14112 \ \Rightarrow & \frac{7 n(n+1)}{2}=14112 \ \Rightarrow & n(n+1)=4032 \ \Rightarrow & n(n+1)=63 	imes 64 \Rightarrow n=63 \end{array} $

If $f:[R \rightarrow[R$ is such that $f(x+y)=f(x)+f(y)$ for all $x, y \in[R . f(\mathrm{l})=7$ and…

Solution