bài 1: \(B = 1 + \frac{9}{45} + \frac{9}{105} + \frac{9}{189} + \dots + \frac{9}{29997}\)

\(B = 1 + \frac{3}{15} + \frac{3}{35} + \frac{3}{63} + \dots + \frac{3}{9999}\)

\(B = 1 + \frac{3}{3 \cdot 5} + \frac{3}{5 \cdot 7} + \frac{3}{7 \cdot 9} + \dots + \frac{3}{99 \cdot 101}\)

\(B = 1 + \frac{3}{2} \cdot \left(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \dots + \frac{1}{99} - \frac{1}{101}\right)\)

\(B = 1 + \frac{3}{2} \cdot \left(\frac{1}{3} - \frac{1}{101}\right)\)

\(B = \frac{150}{101}\)

bài 2:

\(B = 3^1 - 3^2 + 3^3 - 3^4 + \dots + 3^{2023} - 3^{2024}\)

\(3B = 3^2 - 3^3 + 3^4 - 3^5 + \dots + 3^{2024} - 3^{2025}\)

\(3B + B = (3^2 - 3^3 + 3^4 - \dots - 3^{2025}) + (3^1 - 3^2 + 3^3 - \dots - 3^{2024})\)

\(4B=3-3^{2025}\Rightarrow B=\frac{3 - 3^{2025}}{4}\)

bài 3:

\(P = \frac{2 \cdot 8^4 \cdot 27^2 + 4 \cdot 6^9}{2^7 \cdot 6^7 + 2^7 \cdot 40 \cdot 9^4}\)

\(P = \frac{2 \cdot (2^3)^4 \cdot (3^3)^2 + 2^2 \cdot (2 \cdot 3)^9}{2^7 \cdot (2 \cdot 3)^7 + 2^7 \cdot (2^3 \cdot 5) \cdot (3^2)^4}\)

\(P = \frac{2 \cdot 2^{12} \cdot 3^6 + 2^2 \cdot 2^9 \cdot 3^9}{2^7 \cdot 2^7 \cdot 3^7 + 2^7 \cdot 2^3 \cdot 5 \cdot 3^8}\)

\(P = \frac{2^{13} \cdot 3^6 + 2^{11} \cdot 3^9}{2^{14} \cdot 3^7 + 2^{10} \cdot 5 \cdot 3^8}\)

\(P = \frac{2^{11} \cdot 3^6 \cdot (2^2 + 3^3)}{2^{10} \cdot 3^7 \cdot (2^4 + 5 \cdot 3)}\)

\(P=\frac{2^{11} \cdot3^6 \cdot31}{2^{10} \cdot3^7 \cdot31}\)

\(=\frac{2^{11} \cdot3^6}{2^{10} \cdot3^7}=\frac23\)

bài 4: \(S = \frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + \dots + \frac{2014}{4^{2014}}\)

\(\Rightarrow4S=1+\frac{2}{4}+\frac{3}{4^2}+\dots+\frac{2014}{4^{2013}}\)

\(\Rightarrow4S-S=\left(1+\frac{2}{4}+\frac{3}{4^2}+\dots+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\dots+\frac{2014}{4^{2014}}\right)\)

\(\Rightarrow3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+\dots+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)

\(\Rightarrow3S=1+\frac{1}{4}+\frac{1}{4^2}+\dots+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\) (1)

đặt A = \(1+\frac{1}{4}+\frac{1}{4^2}+\dots+\frac{1}{4^{2013}}\) , suy ra

\(4A = 4 + 1 + \frac{1}{4} + \dots + \frac{1}{4^{2012}}\)

\(4A - A = 4 - \frac{1}{4^{2013}}\)
\(\Rightarrow3A=4-\frac{1}{4^{2013}}\)

\(\Rightarrow A=\frac{4}{3}-\frac{1}{3 \cdot4^{2013}}\) (2)

thay (2) vào (1) ta được:

\(3S=\frac{4}{3}-\frac{1}{3 \cdot4^{2013}}-\frac{2014}{4^{2014}}=\frac43-\left(\frac{1}{3 \cdot4^{2013}}+\frac{2014}{4^{2014}}\right)\)

vì \(\left(\frac{1}{3 \cdot4^{2013}}+\frac{2014}{4^{2014}}\right)>0\) nên \(3S<\frac{4}{3}\Rightarrow S<\frac{4}{9}<\frac12\)

bài 5:

ta có: \(50=2\cdot5^2\)

\(75=3\cdot5^2\)

=> \(50^{107}=\left(2\cdot5^2\right)^{107}=2^{107}\cdot5^{214}\)

\(75^73=\left(3\cdot5^2\right)^{73}=3^{73}\cdot5^{146}\)

lập tỉ số để bt số nào lớn hơn ta có:

\(\frac{50^{107}}{75^{73}}=\frac{2^{107}\cdot5^{214}}{3^{73}\cdot5^{146}}\)

\(\frac{50^{107}}{75^{73}}=\frac{2^{107}\cdot5^{68}}{3^{73}}\)

mà ta có \(2^{107}=2^{73}\cdot2^{34}\)

=> tử số= \(2^{107}\cdot5^{68}=2^{73}\cdot2^{34}\cdot5^{68}\)

vì \(2^{73}<3^{73}\)

nhưng \(2^{34}\cdot5^{68}=2^{34}\cdot\left(5^2\right)^{34}=2^{34}\cdot25^{34}=\left(2\cdot25\right)^{34}=50^{34}\)

mà \(50^{34}>27^{34}=\left(3^3\right)^{34}=3^{102}>3^{73}\)

=> \(\frac{50^{107}}{75^{73}}>1\)

=> \(50^{107}>75^{73}\)