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

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

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

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

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

\(\Rightarrowđpcm\)

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

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

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

\(\Rightarrow A=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{\text{4!}}+...+\frac{1}{100!}\right)\)

\(\Rightarrow A=1+1-\frac{1}{99!}-\frac{1}{100!}\)

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

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

\(\Rightarrow A< 2\left(đpcm\right).\)

Chúc bạn học tốt!

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

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

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

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

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

Vậy: A < 1
\(\frac{1}{2}\)

B= \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

= 2. \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

= 2. \(\left(\frac{1}{1}-\frac{1}{100}\right)\) = \(\frac{99}{50}\)

\(\Rightarrow\) B = \(\frac{99}{50}\) < \(\frac{100}{50}\) = 2

Vậy: B < 2

Ta thấy mỗi hạng tử của tổng đều có dạng:  \(\frac{\left(n-1\right)n-1}{n!}=\frac{\left(n-1\right)n}{n!}-\frac{1}{n!}=\frac{1}{\left(n-2\right)!}-\frac{1}{n!}\)

Như vậy VT = \(\frac{1}{0!}-\frac{1}{2!}+\frac{1}{1!}-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+\frac{1}{3!}-\frac{1}{5!}+...+\frac{1}{98!}-\frac{1}{100!}\)

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

LA 0 DO CON NGU DU

bạn đừng nói thế chứ

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 

Chứng minh rằng:

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

Ta có:

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

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

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

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

a ms học xong bài dấu chấm than nè

cho mình hỏi tại sao lại có đoạn 1+1+1/2! ở đâu vậy

1.

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

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

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

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

2.

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

Ta có:

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

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

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

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

Ta có:

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

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

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

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

\(=1+1-\dfrac{1}{99!}-\dfrac{1}{100!}\)

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

Vậy \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}< 2\) (Đpcm)