If f (x) is a function satisfying f(x+y)=f(x).f(y) for all x,yϵ N such that f(1) = 3 and ∑nx=1f(x)=120. Then,the value of n is
A
4
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B
5
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C
6
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D
None of these
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Solution
The correct option is A 4 Given, f(x+y) = f(x) f(y) for all x,yϵN ∴ForanyxϵN,f(x)=[f(1)]x=3x[∴f(1)=3] Since, ∑nx=1f(x)=120 ⇒31+32+33+.....+3n=120 ⇒3n−1=80 ⇒3n=81=34⇒n=4