lon hon 1 nha ban

sửa lại đề : Chứng tỏ rằng : A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)

bài làm

A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}\)

A = \(\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2014-1}{2014!}\)

A = \(1-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2014}{2014!}-\frac{1}{2014!}\)

A = \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2013!}-\frac{1}{2014!}\)

A = \(1-\frac{1}{2014!}< 1\)

lớn hơn 1 nha.

\(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1.2}\)

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

.......................................

\(\frac{1}{2014^2}=\frac{1}{2014\cdot2014}< \frac{1}{2013\cdot2014}\)

\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2013\cdot2014}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)

\(\Rightarrow A< 1-\frac{1}{2014}=\frac{2013}{2014}\)

Trần Nhật Dương    Chứng minh \(A< \frac{3}{4}\) mà :)) 

Ta có: \(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

      ........................

\(\frac{1}{2014^2}=\frac{1}{2014.2014}< \frac{1}{2013.2014}\)

 \(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)

\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)

\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2014}\)

\(\Leftrightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)

\(\Leftrightarrow A< \frac{3}{4}-\frac{1}{2014}< \frac{3}{4}\)

Đề thi đó

có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 +  2^10]

Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]

Q = 2 . 3+2^3 .3 +... + 2^9 .3

Q = 3. [ 2 + 2^3 +... + 2^9]

Vậy Q chia hết cho 3

Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)

\(=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\right)\)

Nhận xét: \(\frac{1}{2^2}=\frac{1}{4}\)

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

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

\(\frac{1}{5^2}< \frac{1}{4\cdot5}\)

...

\(\frac{1}{2014^2}< \frac{1}{2013\cdot2014}\)

Do đó: \(\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\right)< \frac{1}{4}+\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2013\cdot2014}\right)\)

\(\Leftrightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)

\(\Leftrightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)

\(\Leftrightarrow A< \frac{3019}{4028}\)

mà \(\frac{3019}{4028}< \frac{3021}{4028}=\frac{3}{4}\)

nên \(A< \frac{3}{4}\)(đpcm)

Ta có:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2014^2}\)

\(< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2013\cdot2014}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{2013}-\frac{1}{2014}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)

\(=\frac{3}{4}-\frac{1}{2014}\)

\(< \frac{3}{4}\)

Ta có : 

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

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

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

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(A=2.\frac{49}{100}\)

\(A=\frac{49}{50}\)

Lại có : 

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

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

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

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

\(\frac{1}{49^2}>\frac{1}{49.50}\)

\(\Rightarrow\)\(B=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{49^2}>1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(B>1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(B>1+\frac{1}{2}-\frac{1}{50}\)

\(B>1+\frac{12}{25}=\frac{37}{25}=\frac{74}{50}>\frac{49}{50}=A\)

\(\Rightarrow\)\(B>A\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

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

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

........

\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)

=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

Đpcm 

b)B=1/4(1/2^2+1/3^2+...+1/n^2)=1/4*A<1/4

tìm 2 số 3:

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