Question

Consider a function $f(x+y) = f(x)f(y)$ where $x,y$ are positive integers, and $f(1) = 2$. If $f(a+1) + f(a+2) + \dots + f(a+n) = 16(2^n - 1)$, then $a$ is equal to

Accepted Answer

3. $f(x+y) = f(x)f(y)$ with $f(1) = 2 \implies f(x) = 2^x$. $f(a+k) = 2^{a+k} = 2^a \cdot 2^k$. Sum $= 2^a(2^1 + 2^2 + \dots + 2^n) = 2^a \cdot 2(2^n - 1) = 2^{a+1}(2^n - 1)$. Given sum $= 16(2^n - 1) = 2^4(2^n - 1)$. $2^{a+1} = 2^4 \implies a + 1 = 4 \implies a = 3$.

Exponential Function Equation and Summation

Explanation

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