Cho H = 7/3 + 13/32 + 19/33 + . . . + 601/3100.
CMR: 3\(\frac{7}{9}\)< H <5
Những câu hỏi liên quan
Cho \(H=\frac{7}{3}+\frac{13}{3^2}+\frac{19}{3^3}+...+\frac{601}{3^{100}}.\)
Chứng minh:\(3\frac{7}{9}
Cho H = \(\dfrac{7}{3}+\dfrac{13}{3^2}+\dfrac{19}{3^3}+...+\dfrac{601}{3^{100}}\).Chứng minh : \(3\dfrac{7}{9}< H< 5\)
cho H= 7/3+13/3^2+19/3^3+...+601/ 3^100. chứng minh h<5
Cho H = \(\frac{7}{3}+\frac{13}{3^2}+\frac{19}{3^3}+.....+\frac{605}{3^{100}}\)
CMR \(3\frac{7}{9}< H< 5\)
Ta có \(H=\frac{7}{3}+\frac{13}{3^2}+...+\frac{605}{3^{100}}\)
\(\Leftrightarrow3H=7+\frac{13}{3}+...+\frac{605}{3^{99}}\)
\(\Rightarrow2H=7+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2H=7+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Mà \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)
\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2H< 10\)
\(\Leftrightarrow H< 5\left(1\right)\)
Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Mà\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)
hay\(\frac{1}{3^{99}}+\frac{605}{3^{98}}< \frac{22}{9}\)
\(\Rightarrow2H>10-\frac{22}{9}=\frac{68}{9}=2\cdot\left(3+\frac{7}{9}\right)\)
\(\Rightarrow H>3+\frac{7}{9}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)
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\(H=\frac{7}{3}+\frac{13}{3^2^{ }}+\frac{23}{3^3}+...+\frac{601}{3^{100}}\text{
chứng minh }3\frac{7}{9}<H<5
H=\(\frac{7}{3}\)+\(\frac{13}{3^2}\)+...+\(\frac{601}{3^{100}}\)
chứng minh\(3\frac{7}{9}\) <H<5
Chứng minh rằng:
A = 1/3 + 1/32 + 1/33 + ..........+ 1/399 < 1/2
B = 3/12x 22 + 5/22 x 32 + 7/32 x 42 +............+ 19/92 x 102 < 1
C = 1/3 + 2/32 + 3/33 + 4/34 +.........+ 100/3100 ≤ 0
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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Solution
We have: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100)
3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101
Inferred: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
So A = 3101−12
Please help me
Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12
Mà đoạn 2A sai nhé bạn, sửa lại:
2A = 3101−13101−1 2A=-10001
A=-10001/2
A=-5000,5
Vậy A=-5000,5
Giúp tôi với mọi người ơi:Cho C1/2x3/4x5/6x......x2017/2018CMR:C^21/2019Bài típ nè:Cho D1/2x3/4x.......x99/100CMR:1/15D1/10Nữa nè:Cho H7/3+13/3^2+19/3^3+......+601/3^100CMR:34/9H5Nhớ giải rõ ràng nha! Thanks mọi người!
Đọc tiếp
Giúp tôi với mọi người ơi:
Cho C=1/2x3/4x5/6x......x2017/2018
CMR:C^2<1/2019
Bài típ nè:
Cho D=1/2x3/4x.......x99/100
CMR:1/15<D<1/10
Nữa nè:
Cho H=7/3+13/3^2+19/3^3+......+601/3^100
CMR:34/9<H<5
Nhớ giải rõ ràng nha! Thanks mọi người!