Let N be the set of natural numbers. Suppose f:N→N is a function satisfying the following conditions (a)f(m+n)=f(m)+f(n)(b)f(2)=2
The value of 1720∑k=1f(k) is
A
30.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
30.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
30
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
f:N→N f(m+n)=f(m)+f(n)
Putting m=n=1, we get f(2)=2f(1)=2⇒f(1)=1
Putting m=2,n=1, we get f(3)=f(1)+f(2)=3
Putting m=n=2, we get f(4)=2f(2)=4