Question

If f : R → R satisfies f(x+y) = f(x) + f(y), for all x, y∈R and f(1) =7, then $\sum \limits^{n}_{r=1}$, f(r) is
(a) $\frac{7n(n+1)}{2}$
(b) $\frac{7n}{2}$
(c) $\frac{7(n+1)}{2}$
(d) 7n + (n +1) .

Answer 1

Q).

If f is a function satisfying f(x+y)=f(x).f(y)∀x,y∈N such that f(1)=3and∑x=1nf(x)=120, then find the value of n.

3456

Sum of n terms of G.P. =Sn=arn−1r−1

Given: f(x+y)=f(x).f(y),

∑nx=1f(x)=120 and f(1)=3.

when x=y=1, f(1+1)=f(2)=f(1).f(1)=3×3=9

when x=2andy=1,f(2+1)=f(3)=f(2).f(1)=9×3=27

Now

Given: ∑nx=1f(x)==f(1)+f(2)+f(3)+........f(n)=120

⇒3+9+27+........f(n)=120

This series is a G.P. with a=3,r=3andSn=120

We know that Sum of n terms of G.P. =Sn=arn−1r−1

⇒3.3n−13−1=120

⇒3n−1=40×2=80

⇒3n=81=34 ⇒n=4.

i have solved everything... i hope it help you