The value of $\frac{2^{n-1} - 2^n}{2^{n+4} + 2^{n+1}}$ is

### Concept & Formula The core concept is factoring exponential expressions by extracting the common term with the lowest power to simplify the numerator and denominator separately. The primary rule utilized is: $$ a^{m+n} = a^m \times a^n $$ ### Step-by-Step Solution * **Step 1:** Isolate the lowest power of 2 in both the numerator and the denominator to factor them out. In the numerator ($2^{n-1} - 2^n$), the lowest power is $n-1$. Note that $2^n$ can be written as $2^{n-1} \times 2^1$. In the denominator ($2^{n+4} + 2^{n+1}$), the lowest power is $n+1$. Note that $2^{n+4}$ can be written as $2^{n+1} \times 2^3$. * **Step 2:** Factor out these lowest power terms. Numerator: $2^{n-1} (1 - 2^1) = 2^{n-1} (1 - 2) = 2^{n-1} (-1) = -2^{n-1}$ Denominator: $2^{n+1} (2^3 + 1) = 2^{n+1} (8 + 1) = 2^{n+1} (9)$ * **Step 3:** Reconstruct the fraction with the simplified terms. $\frac{-2^{n-1}}{9 \times 2^{n+1}}$ * **Step 4:** Simplify the exponent parts by bringing $2^{n-1}$ down to the denominator using the division law $a^m \div a^n = a^{m-n}$. $\frac{-1}{9 \times 2^{(n+1) - (n-1)}}$ $\frac{-1}{9 \times 2^{n+1-n+1}}$ $\frac{-1}{9 \times 2^2}$ * **Step 5:** Calculate the final arithmetic value. $\frac{-1}{9 \times 4} = -\frac{1}{36}$ ### Exam Strategy & Shortcut **Assume $n=1$**: Since the options are all distinct numerical constants, the outcome is independent of the variable $n$. Choosing a small value like $n=1$ makes the calculation trivial. Numerator: $2^0 - 2^1 = 1 - 2 = -1$ Denominator: $2^5 + 2^2 = 32 + 4 = 36$ Result: $-\frac{1}{36}$. This takes only moments to execute mentally. ### Common Pitfall A very common algebraic mistake here is mishandling the subtraction in the numerator ($1 - 2$). Students frequently drop the negative sign by mistake and answer $\frac{1}{36}$, which isn't an option here but could easily trap someone if it were. ### Final Answer **Therefore, the correct answer is $-\frac{1}{36}$.**