1. **Stating the problem:** Simplify the expression $$\frac{25^n - 1 \cdot 6^n}{10^{n-1} \cdot 5^n}$$. 2. **Recall the rules:** - Powers with the same base: $$a^m \cdot a^n = a^{m+n}$$. - Division of powers with the same base: $$\frac{a^m}{a^n} = a^{m-n}$$. - Express numbers as prime factors to simplify. 3. **Rewrite bases in prime factors:** - $$25 = 5^2$$ - $$6 = 2 \cdot 3$$ - $$10 = 2 \cdot 5$$ 4. **Rewrite numerator:** $$25^n - 1 \cdot 6^n = 5^{2n} - 6^n$$ 5. **Rewrite denominator:** $$10^{n-1} \cdot 5^n = (2 \cdot 5)^{n-1} \cdot 5^n = 2^{n-1} \cdot 5^{n-1} \cdot 5^n = 2^{n-1} \cdot 5^{2n-1}$$ 6. **Put together:** $$\frac{5^{2n} - 6^n}{2^{n-1} \cdot 5^{2n-1}} = \frac{5^{2n}}{2^{n-1} \cdot 5^{2n-1}} - \frac{6^n}{2^{n-1} \cdot 5^{2n-1}}$$ 7. **Simplify first term:** $$\frac{5^{2n}}{5^{2n-1}} = 5^{2n - (2n-1)} = 5^1 = 5$$ So first term is $$\frac{5}{2^{n-1}}$$. 8. **Simplify second term:** Rewrite $$6^n = (2 \cdot 3)^n = 2^n \cdot 3^n$$ So second term is $$\frac{2^n \cdot 3^n}{2^{n-1} \cdot 5^{2n-1}} = \frac{2^n}{2^{n-1}} \cdot \frac{3^n}{5^{2n-1}} = 2^{n-(n-1)} \cdot \frac{3^n}{5^{2n-1}} = 2 \cdot \frac{3^n}{5^{2n-1}}$$ 9. **Final expression:** $$\frac{25^n - 1 \cdot 6^n}{10^{n-1} \cdot 5^n} = \frac{5}{2^{n-1}} - 2 \cdot \frac{3^n}{5^{2n-1}}$$ This is the simplified form. **Final answer:** $$\boxed{\frac{5}{2^{n-1}} - 2 \cdot \frac{3^n}{5^{2n-1}}}$$