Relations & Functions Question - None

Question

Let $\sum_{k=1}^{10} f(a+k) = 16(2^{10}-1)$, where the function $f$ satisfies $f(x+y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2$. Then the natural number $a$ is ______.

Accepted Answer

Step 1: Identify the function from the functional equation

Given: f(x+y) = f(x)f(y) and f(1) = 2

This is Cauchy's exponential functional equation. For natural numbers:

f(2) = f(1+1) = f(1)·f(1) = 2·2 = 4 = 2²

f(3) = f(2+1) = f(2)·f(1) = 4·2 = 8 = 2³

Therefore, f(n) = 2^n for all natural numbers n

Step 2: Set up the sum

∑_{k=1}^{10} f(a+k) = ∑_{k=1}^{10} 2^{a+k} = 2^a · ∑_{k=1}^{10} 2^k

Step 3: Evaluate the geometric series

∑_{k=1}^{10} 2^k = 2 + 2² + 2³ + ... + 2^{10}

This is a geometric series with first term 2, ratio 2, and 10 terms:

∑_{k=1}^{10} 2^k = 2·(2^{10} - 1)/(2 - 1) = 2(2^{10} - 1)

Step 4: Solve for a

2^a · 2(2^{10} - 1) = 16(2^{10} - 1)

2^a · 2 = 16

2^{a+1} = 2^4

a + 1 = 4

∴ a = 3

Question:

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