Let f: R arrow R be a function which satisfies f ( x + y )= f ( x )+ f ( y ) ∀ x , y ∈ R . If f (1)=2 and g(n)= displaystyle∑k=1(n-1) f(k), n ∈ N then the value of n, for which g(n)=20, is :

Question

Let f: R arrow R be a function which satisfies f ( x + y )= f ( x )+ f ( y ) ∀ x , y ∈ R . If f (1)=2 and g(n)= displaystyle∑k=1(n-1) f(k), n ∈ N then the value of n, for which g(n)=20, is : (A) 5 (B) 9 (C) 20 (D) 4

Tardigrade Admin · Accepted Answer

f(x+y)=f(x)+f(y) ⇒ f(n)=n f(1) f(n)=2 n g(n)= displaystyle∑k=1n-1 2 n=2(((n-1) n/2))=n(n-1) g(n)=20 ⇒ n(n-1)=20 n=5

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

5

9

20

4

$f (x + y) = f (x) + f (y)$
$\Rightarrow f (n) = n f (1)$
$f (n) = 2 n$
$g (n) = k = 1 \sum n - 1 2 n = 2 (\frac{( n - 1 ) n}{2}) = n (n - 1)$
$g (n) = 20 \Rightarrow n (n - 1) = 20$
$n = 5$

Q. Let f:R→R be a function which satisfies f(x+y)=f(x)+f(y)∀x,y∈R. If f(1)=2 and g(n)=k=1∑(n−1)​f(k),n∈N then the value of n, for which g(n)=20, is :

5

9

20

4

f(x+y)=f(x)+f(y) ⇒f(n)=nf(1) f(n)=2n g(n)=k=1∑n−1​2n=2(2(n−1)n​)=n(n−1) g(n)=20⇒n(n−1)=20 n=5

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

$f (x + y) = f (x) + f (y)$
$\Rightarrow f (n) = n f (1)$
$f (n) = 2 n$
$g (n) = k = 1 \sum n - 1 2 n = 2 (\frac{( n - 1 ) n}{2}) = n (n - 1)$
$g (n) = 20 \Rightarrow n (n - 1) = 20$
$n = 5$