Question

Q1.f(x) satisfies f(x+y)=f(x)+f(y) for all x,y belongs to R , f(1)=5 then find
m
singma [f(n)]
n=1

Answer 1

Explanation:

Given,

f(x) satisfies f(x+y) = f(x) + f(y) for all x, y belongs to R.

Consider :

f(x) = kx

It satisfies the given equation

⇒f(x+y) = k(x+y) = kx + ky

⇒f(x) = kx

⇒f(y) = ky

So f(x) + f(y) = f(x+y)

According to the question,

f(1) = 5

⇒ k(1) = 5

⇒ k = 5

Therefore, The function f(x) = 5x

To find :

$\begin{gathered} \sum \limits_{n = 1}^{m} f(n) \\ \\ = \sum \limits_{n = 1}^{m} \: 5n \\ \\ = 5 \sum \limits_{n = 1}^{m} n \\ \\ = 5 \frac{m(m + 1)}{2} \\ \\ = 5m( \frac{m + 1}{2} )\end{gathered}$

Therefore, The required answer is 5m(m+1)/2.