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if f(x+y)=f(x)+f(y) and f(1)=7 then ∑f(r) from r=1 to r=n?

if f(x+y)=f(x)+f(y) and f(1)=7 then ∑f(r) from r=1 to r=n?

Grade:12th pass

4 Answers

Arun
25750 Points
4 years ago
Dear student
 
if f(x+y)=f(x)+f(y)
Then f(x) = e^(kx)
 
Hence e^k = 7
k = ln 7
 
Hence f(x) = e(ln7)x = 7x
 
Hence from r = 1 to n
 
sigma f(r) = 7 + 7² + 7³ + 7⁴ +.....7ⁿ
 
 = 7 ( 7ⁿ - 1) /(7-1)
 
 = 7(7ⁿ - 1)/6
Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
Dear student

f(1) = 7
f(2) = 2* f(1) = 2*7
f(3) = f(2)+f(1) = 2*7 +7
...
f(n) = n* f(1) = 7*n

Now sum can be obtained

Regards

Aditya Gupta
2081 Points
4 years ago
Note that f(r+1)= f(r)+f(1)
Or f(r+1)-f(r)= f(1), Which is an AP for r belonging to natural numbers.
Hence, f(r)= f(1)+(r-1)*f(1)
Or f(r)= r*f(1)= 7r
Hence summation from 1 to n is
S= 7n(n+1)/2
 
Vikas TU
14149 Points
4 years ago
Dear student 
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.
 

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