1. We are given the expression $$\frac{2^n}{3 - 2^{-n}}$$ and asked to find its smallest term. 2. First, rewrite the expression for clarity: $$f(n) = \frac{2^n}{3 - \frac{1}{2^n}} = \frac{2^n}{3 - 2^{-n}}$$ 3. To analyze this expression, consider the denominator: $$3 - 2^{-n}$$ Since $$2^{-n} = \frac{1}{2^n} > 0$$, the denominator is smaller when $$2^{-n}$$ is larger. 4. We want to find the smallest term, so we consider the behavior of $$f(n)$$ as $$n$$ varies over integers, commonly all integers or non-negative integers. 5. Rewrite $$f(n)$$ by multiplying numerator and denominator by $$2^n$$ to avoid negative exponents: $$f(n) = \frac{2^n}{3 - 2^{-n}} \times \frac{2^n}{2^n} = \frac{2^{2n}}{3\cdot 2^n - 1}$$ 6. Now, consider $$f(n) = \frac{2^{2n}}{3 \cdot 2^n - 1}$$. 7. Let $$x = 2^n > 0$$, then $$f(x) = \frac{x^2}{3x - 1}$$ where $$x > 0$$ and $$3x -1 \neq 0 \Rightarrow x \neq \frac{1}{3}$$. 8. To find extrema, take the derivative: $$f'(x) = \frac{(2x)(3x -1) - x^2(3)}{(3x -1)^2} = \frac{2x(3x -1) - 3x^2}{(3x -1)^2}$$ $$= \frac{6x^2 - 2x - 3x^2}{(3x -1)^2} = \frac{3x^2 - 2x}{(3x -1)^2}$$ 9. Set numerator equal to zero for critical points: $$3x^2 - 2x = 0 \Rightarrow x(3x - 2) = 0$$ Solutions: $$x=0$$ (not valid since $$x=2^n>0$$) and $$x=\frac{2}{3}$$. 10. Evaluate $$f(x)$$ at $$x=\frac{2}{3}$$: $$f\left(\frac{2}{3}\right) = \frac{(\frac{2}{3})^2}{3\cdot \frac{2}{3} -1} = \frac{\frac{4}{9}}{2 - 1} = \frac{4}{9}$$. 11. Since $$f(x)$$ tends to infinity as $$x\to \frac{1}{3}^+$$ and also grows large as $$x\to \infty$$, the minimum value is at $$x=\frac{2}{3}$$. 12. Recall $$x=2^n$$, so solve for $$n$$: $$2^n = \frac{2}{3}$$ Taking log base 2, $$n = \log_2 \left(\frac{2}{3}\right) = 1 - \log_2 3 \approx 1 - 1.58496 = -0.58496$$ Since $$n$$ is not necessarily integer, for integer values check neighboring values. 13. For integer $$n = -1$$: $$f(-1) = \frac{2^{-1}}{3 - 2^{1}} = \frac{\frac{1}{2}}{3 - 2} = \frac{1/2}{1} = \frac{1}{2} = 0.5$$ For $$n=0$$: $$f(0) = \frac{1}{3 - 1} = \frac{1}{2} = 0.5$$ For $$n = -2$$: $$f(-2) = \frac{2^{-2}}{3 - 2^2} = \frac{1/4}{3 - 4} = \frac{1/4}{-1} = -\frac{1}{4} = -0.25$$ But denominator is negative at $$n=-2$$, so value is negative. 14. Since the function is defined for $$3 - 2^{-n} \neq 0$$, negative values can occur. 15. For values $$n < -1.58496$$, denominator becomes negative and function changes sign. 16. Considering domain and behavior, the smallest term of the expression is not bounded since for $$n$$ making denominator close to zero from the right side, the function tends to infinity. 17. But for integer $$n$$, checking values $$n=-2$$ gives $$-0.25$$, which is less than $$f(-1)=0.5$$ and $$f(0)=0.5$$. For $$n = -3$$: $$f(-3) = \frac{2^{-3}}{3 - 2^3} = \frac{1/8}{3 - 8} = \frac{1/8}{-5} = -\frac{1}{40} = -0.025$$ '> -0.25$$, so minimum is at $$n=-2$$ among integer values. **Final Answer:** The smallest term of the expression $$\frac{2^n}{3 - 2^{-n}}$$ for integer $$n$$ is $$-\frac{1}{4}$$ at $$n=-2$$.