C=1/3-2/32+3/33-4/34+...+99/399-100/3100<3/16
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H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
27 tháng 9 2021
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
KJ
1
Q
13 tháng 12 2021
Tham khảo
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)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12
KJ
2
16 tháng 12 2021
\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)
\(3C=3.\left(\frac{1}{3}-\frac{2}{3^2}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\) )
\(\Rightarrow3C=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3C+C=4C=1-\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
Đặt A=\(1-\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3A=3\times\left(1-\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
\(\Rightarrow3A=3-1+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow3A+A=4A=3-\frac{1}{3^{99}}\)
\(\Rightarrow A=\frac{1}{4}\times\left(3-\frac{1}{3^{99}}\right)\)
Thay A vào ta có : \(4C=\frac{1}{4}\times\left(3-\frac{1}{3^{99}}\right)-\frac{1}{3^{100}}\)
\(C=\frac{1}{16}\times\left(3-\frac{1}{3^{99}}\right)-\frac{25}{3^{100}}\)
\(C=\frac{3}{16}-\frac{1}{16\times3^{99}}-\frac{25}{3^{100}}< \frac{3}{16}\)
Vậy C <\(\frac{3}{16}\)
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