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Question
If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and `sum_(x = 1)^n` f(x) = 120, find the value of n.
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Solution
It is given that,
f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)
f (1) = 3
Taking x = y = 1 in (1), we obtain
f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9
Similarly,
f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27
f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81
∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.
It is known that, Sn = `(a(r^n - 1))/(r - 1)`
It is given that, `sum_(x = 1)^nf (x) = 120`
∴ `120 = (3(3^n - 1))/(3 - 1)`
= `120 = 3/2 (3^n - 1)`
= 3n - 1 = 80
= 3n - 1 = 81 = 34
∴ Thus, the value of n is 4.
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