Tính: 100 + 97 + 94 + ... + 4 + 1.

Lời giải:

\[\frac{1}{9} \times {27^n} = {3^n}\]

\[\frac{1}{9} = \frac{{{3^n}}}{{{{27}^n}}}\]

\[\frac{1}{9} = {\left( {\frac{3}{{27}}} \right)^n} = {\left( {\frac{1}{9}} \right)^n}\]

n = 1

Vậy n = 1.

Lời giải:

\[1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + ... + \frac{2}{{x\left( {x + 1} \right)}} = 1 + \frac{{2023}}{{2025}}\]

\[\frac{2}{2} + \frac{2}{6} + ... + \frac{2}{{x\left( {x + 1} \right)}} = \frac{{4048}}{{2025}}\]

\[\frac{1}{2} + \frac{1}{6} + ... + \frac{1}{{x\left( {x + 1} \right)}} = \frac{{2024}}{{2025}}\]

\[\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + ... + \frac{1}{{x \cdot \left( {x + 1} \right)}} = \frac{{2024}}{{2025}}\]

\[1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{x} - \frac{1}{{x + 1}} = \frac{{2024}}{{2025}}\]

\[1 - \frac{1}{{x + 1}} = \frac{{2024}}{{2025}}\]

\[\frac{1}{{x + 1}} = \frac{1}{{2025}}\]

x + 1 = 2025

x = 2025 ‒ 1

x = 2024

Vậy x = 2024.