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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!

1.\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Thấy:\(\frac{1}{2^{100}}>0\Rightarrow1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow A< 1\)

Ta có:\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)=A+100< 1+100=101\)

\(101>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)\ge100\)

\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(\frac{1}{2^{100}}\right)>\left(\frac{101}{100}\right)^{100}>3\)

*Cách khác:

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)

\(=\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)

Ta thấy:

\(\frac{2+1}{2}>\frac{2^2+1}{2^2}>....>\frac{2^{100}+1}{2^{100}}\)

\(\Rightarrow\frac{2+1}{2}>\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)

Mà \(\frac{2+1}{2}< 3\)

\(\Rightarrow\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}< 3\)

\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)< 3\)

=A*(1/100-1/10^2)

=0

khó vậy 

\(=\left(\dfrac{1}{100}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{10^2}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=0\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)=0\)

Ta chia thành hai vế (1) và (2)

Số số hạng (1) là :

( 101 - 1 ) : 1 + 1 = 101  ( số )

Tổng (1) là :

( 101 + 1 ) x 101 : 2 = 5151

Tự tính tiếp

\(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99+100\right)\)

\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+...+3\right)+...+\left(99+99\right)+100\)

\(=1.100+2.99+3.98+...+99.2+100.1\)

Do đó kết quả của phép tính cần tìm là: 

\(\frac{1.100+2.99+...+99.2+100.1}{\left(1.100+2.99+...+99.2+100.1\right).2013}=\frac{1}{2013}\)

bạn tính cái tổng trong ngoặc ra theo công thức là ra