If the function fsatisfies the relation f(x+y)=f(x)⋅f(y) for all natural numbers x,y, f(1)=2 and ∑r=1n f(a+r)=162n−1 ,then the natural number α is

f(2)=f(1+1)=f(1)×f(1)=22f(3)=f(2+1)=f(2)×f(1)=22×2=23∴f(n)=2n⇒162n−1=∑r=1n f(a+r)=∑r=1n 2a+r=2a2+22+…+2n=2a⋅22n−12−1=2a+12n−1 ∴ 2a+1=16 or a=3

If the function fsatisfies the relation f(x+y)=f(x)⋅f(y) for all natural numbers x,y, f(1)=2 and ∑r=1n f(a+r)=162n−1 ,then the natural number α is

f(2)=f(1+1)=f(1)×f(1)=22f(3)=f(2+1)=f(2)×f(1)=22×2=23∴f(n)=2n⇒162n−1=∑r=1n f(a+r)=∑r=1n 2a+r=2a2+22+…+2n=2a⋅22n−12−1=2a+12n−1 ∴ 2a+1=16 or a=3

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If the function $f$ satisfies the relation $f (x + y) = f (x) \cdot f (y)$ for all natural numbers $x, y,$ $f (1) = 2 and \sum_{r = 1}^{n} f (a + r) = 16 (2^{n} - 1)$ ,then the natural number $α$ is

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$f (2) = f (1 + 1) = f (1) \times f (1) = 2^{2}$

$f (3) = f (2 + 1) = f (2) \times f (1) = 2^{2} \times 2 = 2^{3}$

$\begin{aligned} ∴ f (n) = 2^{n} \\ \Rightarrow 16 (2^{n} - 1) = \sum_{r = 1}^{n} f (a + r) = \sum_{r = 1}^{n} 2^{a + r} \end{aligned}$

$\begin{aligned} = 2^{a} (2 + 2^{2} + \dots + 2^{n}) = 2^{a} \cdot 2 (\frac{2^{n} - 1}{2 - 1}) = 2^{a + 1} (2^{n} - 1) \end{aligned}$

$∴ 2^{a + 1} = 16 or a = 3$

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If the function fsatisfies the relation f(x+y)=f(x)⋅f(y) for all natural numbers x,y, f(1)=2 and ∑r=1n f(a+r)=162n−1 ,then the natural number α is

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If the function $f$ satisfies the relation $f (x + y) = f (x) \cdot f (y)$ for all natural numbers $x, y,$ $f (1) = 2 and \sum_{r = 1}^{n} f (a + r) = 16 (2^{n} - 1)$ ,then the natural number $α$ is