Consider a function $f$ satisfying $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \mathrm{f}(\mathrm{y})$ where $x , y$ are positive integers, and $f(1)=2$. If $f(\mathrm{a}+1)+f(\mathrm{a}+2)+\ldots \ldots+f(\mathrm{a}+\mathrm{n})=$ $16\left(2^{n}-1\right)$ then a is equal to

We're given that $f(x+y) = f(x) \cdot f(y)$ where $x, y$ are positive integers, and $f(1) = 2$.

This type of equation is called a multiplicative functional equation. It means that when we add inputs, we multiply outputs.

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Starting with $f(1) = 2$:

$f(2) = f(1+1) = f(1) \cdot f(1) = 2 \times 2 = 4$

$f(3) = f(2+1) = f(2) \cdot f(1) = 4 \times 2 = 8$

$f(4) = f(3+1) = f(3) \cdot f(1) = 8 \times 2 = 16$

The pattern emerges:

$f(1) = 2 = 2^1$

$f(2) = 4 = 2^2$

$f(3) = 8 = 2^3$

$f(4) = 16 = 2^4$

Therefore: $f(x) = 2^x$

We can verify: $f(x+y) = 2^{x+y} = 2^x \cdot 2^y = f(x) \cdot f(y)$

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Using the given condition:

$f(a+1) + f(a+2) + f(a+3) + \ldots + f(a+n) = 16(2^n - 1)$

Substituting $f(x) = 2^x$:

$2^{a+1} + 2^{a+2} + 2^{a+3} + \ldots + 2^{a+n} = 16(2^n - 1)$

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The left side is a geometric series. Factoring out $2^{a+1}$:

$2^{a+1}(1 + 2 + 2^2 + \ldots + 2^{n-1}) = 16(2^n - 1)$

The expression in parentheses is a geometric series with first term = 1, common ratio = 2, and number of terms = n.

The sum of this geometric series is: $\dfrac{2^n - 1}{2 - 1} = 2^n - 1$

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$2^{a+1} \cdot (2^n - 1) = 16(2^n - 1)$

Since $(2^n - 1) \neq 0$, we can divide both sides by $(2^n - 1)$:

$2^{a+1} = 16$

$2^{a+1} = 2^4$

Therefore: $a + 1 = 4$

$a = 3$