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f(x + y) = f(x) f(y): sum of f(k) as a G.P.

This type of question, asked 3 times: 25 Jun 2022, 24 Jan 2023, 22 Jan 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    64
    Let f:N→Rf:N \rightarrow R be a function such that f(x+y)=2f(x)f(y)f(x+y) = 2f(x)f(y) for natural numbers xx and yy. If f(1)=2f(1) = 2, then the value of α\alpha for which ∑k=110f(α+k)=5123(220−1)\sum_{k = 1}^{10} f(\alpha+ k) =\frac{512}{3}\left( 2^{20}- 1 \right) holds, is

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  2. Question 2

    24 Jan 2023

    163
    Let f(x)f(x) be a function such that f(x+y)=f(x)⋅f(y)f(x + y) = f(x) \cdot f(y) for all x,y∈Nx,y \in N. If f(1)=3f(1) = 3 and ∑k=1nf(k)=3279\sum_{k = 1}^{n} f(k) = 3279, then the value of nn is

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  3. Question 3

    22 Jan 2026 S2

    69
    Let ff and gg be functions satisfying f(x+y)=f(x)f(y),f(1)=7f(x + y) = f(x)f(y),f(1) = 7 and g(x+y)=g(xy)g(x + y) = g(xy), g(1)=1g(1) = 1, for all x,y∈Nx,y \in N. If ∑x=1n(f(x)g(x))=19607\sum_{x = 1}^{n} \left( \frac{f(x)}{g(x)} \right)= 19607, then n is equal to :

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