a.Chứng tỏ rằng B = 1/2² + 1/3² + 1/4² + 1/5² + 1/6² + 1/7² +1/8² < 1

b.Cho S = 3/1.4 + 3/4.7 + 3/7.10 +......+3/40.43 + 3/43.46 hãy chứng tỏ rằng S < 1

Xin lỗi mọi người mình tính đặt câu hỏi nhưng ấn nhầm phần trả lời ạ!

\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

Đặt N = 1 + 2 + 2² +...+ 2²⁰¹²

2N = 2 + 2² + 2³ +...+ 2²⁰¹³

2N - N = (2 + 2² + 2³+....+ 2²⁰¹³) - (1 + 2 + 2² +....+ 2²⁰¹²)

N = 2²⁰¹³ - 1

Thay N vào M ta được:

Đặt \(N=1+2+2^2+...+2^{2012}\)

\(2N=2+2^2+2^3+...+2^{2013}\)

\(2N-N=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)

\(N=2^{2013}-1\)

Thay N vào M ta được:

\(M=\dfrac{2^{2013-1}}{2^{2014}-2}=\dfrac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\dfrac{1}{2}\)

Ta có : 

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)

\(2A=1+2+\frac{1}{2}+...+\frac{1}{2^{2011}}\)

\(2A-A=\left(1+2+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

\(A=\frac{2^{2013}-1}{2^{2012}}\)

Vậy \(A=\frac{2^{2013}-1}{2^{2012}}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)

=>2A=\(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)

=>2A-A=\(\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)=2-\frac{1}{2^{2012}}\)

=>A=\(\frac{2^{2013}-1}{2^{2012}}\)

\(\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}-1\right)\\ ...\\ 2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\\ 2A=3^{128}-1\)

Vậy \(A=\dfrac{3^{128}-1}{2}.\)