Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)

=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=             \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=  \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)

\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)

cunasai

A = \(\frac{1}{3}\)  + \(\frac{2}{3^2}\)  + \(\frac{3}{3^3}\) + \(\frac{4}{3^4}\) +....+ \(\frac{100}{3^{100}}\) 

3A = 1 + \(\frac{2}{3}\) + \(\frac{3}{3^2}\) + \(\frac{4}{3^3}\)  +...+ \(\frac{100}{3^{99}}\) 

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

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

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

\(\Rightarrow\) 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\) 

\(\Rightarrow\) 2B = \(1-\frac{1}{3^{99}}\)  

\(\Rightarrow\) \(B=\left(1-\frac{1}{3^{99}}\right):2\)   

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

Vậy A < \(\frac{3}{4}\) 

Vậy:...........

Đặt :

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

\(\Leftrightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+....+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+....+\frac{100}{3^{100}}\right)\)

\(\Leftrightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt : \(H=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\) \(\Leftrightarrow2A=H-\frac{100}{3^{100}}\)

\(\Leftrightarrow3H=3+1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(\Leftrightarrow3H-H=\left(4+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Leftrightarrow2H=3-\frac{1}{3^{99}}\)

\(\Leftrightarrow H=\frac{3-\frac{1}{99}}{2}\)

\(\Leftrightarrow2A=\frac{3-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{2.3^{100}}\)

\(\Leftrightarrow A< \frac{3}{4}\left(đpcm\right)\)

Test câu trả lời

Chúc bạn học tốt!

hình như đề sai bởi vì trong dãy số có số 4/4^3

\(\frac{4}{3^4}\)moi dung

đề sai a

Tham khảo nha bạn :

Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến

Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó

\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)

Nên từ đây => \(A< 1\)     (ĐPCM)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)

Cộng vế theo vế

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

=> \(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(A< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)(1)

Lại có \(\frac{99}{100}< 1\)(2)

Từ (1) và (2) => \(A< \frac{99}{100}< 1\Rightarrow A< 1\left(đpcm\right)\)

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sửa đề câu 1 :

\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=1-\frac{1}{100!}< 1\)

sửa đề câu 2

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

khi cộng cac số có tử bé hơn mẫu thì tổng sẽ <1 nha