Question

Let $f:R\to R$ be a function which satisfies $f(x+y)=f\left(x\right)+f\left(y\right)\forall x,y\in R$. If $f\left(1\right)=2$ and $g\left(n\right)=\underset{k=1}{\overset{n-1}{\sum f\left(k\right)}}$, $n\in N$ then the value of $n$ for which $g\left(n\right)=20$is:

• A. $9$
• B. $5$
• C. $4$
• D. $20$

Answer 1

The correct option is B

$5$

Determine the value of $n$ for which $g\left(n\right)=20$

We have,

$f(x+y)=f\left(x\right)+f\left(y\right)\forall x,y\in R$

$g\left(n\right)=\underset{k=1}{\overset{n-1}{\sum f\left(k\right)}}$

We know that: $f\left(1\right)=2$

$$\begin{gathered} Letx=y=1 \\ f(1+1)=f\left(1\right)+f\left(1\right) \\ \Rightarrow f\left(2\right)=2f\left(1\right) \\ \Rightarrow f\left(2\right)=4 \end{gathered}$$

Similarly,

$$\begin{gathered} Letx=1,y=2 \\ f(1+2)=f\left(1\right)+f\left(2\right) \\ \Rightarrow f\left(3\right)=2+4 \\ \Rightarrow f\left(3\right)=6 \end{gathered}$$

$$\begin{gathered} g\left(n\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)+....+f(n-1) \\ =2+4+6+....2(n-1) \\ =2(1+2+3...(n-1\left)\right) \\ =2.\frac{(n-1)n}{2} \\ =n(n-1) \\ =n^2-n \end{gathered}$$

Since we know, $g\left(n\right)=20$, placing the value we get,

$$\begin{gathered} \Rightarrow n^2-n=20 \\ \Rightarrow n^2-n-20=0 \\ \Rightarrow (n-5)(n+4)=0 \\ \Rightarrow n=-4,5\left[ncannotbenegative\right] \\ \therefore n=5 \end{gathered}$$

Therefore, the value of $n$ for which $g\left(n\right)=20$is: $5.$

Hence, the correct answer is Option (B).