Each term of S is n!(n² + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!

By definition, n(n + 1)n! + n! = n! + n(n + 1)!

Therefore, S can be simplified as

1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!

So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)

\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)

\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)

\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)

\(=...\)

\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)

\(=\dfrac{101!+100\cdot101!}{101!}\)

\(=1+100=101\)

Hence, \(\dfrac{S+1}{101!}=101\)

 ->  M = (100 – 1).(100 – 2^2). (100 – 3^2)…(100 – 50^2)

M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2) .(100 – 10^2) .(100 – 11^2) …(100 – 50^2)

M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2). (100 – 100) .(100 – 11^2) …(100 – 50^2)

M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2) .0.(100 – 11^2) …(100 – 50^2)

M = 0

Vậy M = 0.

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

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

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

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

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

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

\(\frac{1}{2^2}>\frac{1}{2.3}\)

\(\frac{1}{3^2}>\frac{1}{3.4}\)

\(\frac{1}{4^2}>\frac{1}{4.5}\)

\(............\)

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow\)\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

\(\Rightarrow\)\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow\)\(A>\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow\)\(A>\frac{99}{202}\) \(\left(1\right)\)

Lại có : 

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(............\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\)\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(\Rightarrow\)\(A< \frac{99}{100}\) \(\left(2\right)\)

Từ (1) và (2) suy ra : \(\frac{99}{202}< A< \frac{99}{100}\) ( đpcm ) 

Vậy \(\frac{99}{202}< A< \frac{99}{100}\)

Chúc bạn học tốt ~ 

câu hỏi là gì

Viết các hữu tỉ sau dưới dạng phân số tối giản có mẫu dương

bn ơi , bn có thể ghi rõ đề bài đc ko??

ghi làm sao vậy bạn

a:

Số số hạng trong dãy M là:

(1002-12):10+1=100(số)

=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10

\(M=1002-992+982-972+...+22-12\)

\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)

\(=10+10+...+10\)

=10*50=500

b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)

\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)

=10+10+...+10

=10*10=100

Ta có:

\(M=\dfrac{100^{100}+1}{100^{99}+1}\)

\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}\)

\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}\)

\(\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}\) 

\(N=\dfrac{100^{101}+1}{100^{100}+1}\)

\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}\)

\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}\)

\(\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}\)

Mà: \(100^{101}>100^{100}\)

\(\Rightarrow100^{101}+100>100^{100}+100\)

\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)

\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)

\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)

\(\Rightarrow N< M\)