Sets, Relations & Functions Question - None

Question

Given $\displaystyle\sum_{k=1}^{10} f(a+k) = 16(2^{10}-1)$, where $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{N}$ and $f(1) = 2$. Find the value of $a$.

Accepted Answer

Step 1: Determine f(n) from the functional equation.

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

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

By induction: f(n) = 2^n

Step 2: Rewrite the given 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 = 2¹ + 2² + ... + 2^{10} = 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 (Answer: B)

Question:

Step-by-Step Solution

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