cho M= 1/3 + 2/32 + 3/33 +.....+ 100/3100. Chững minh M<3/4
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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
VG
1
13 tháng 6 2019
Ta có: 3A = 3.(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−1
⇒ A = 3101−1
2
Vậy A = 3101−1
2
LT
0
BK
1
S
1
S
0
M = 1/3^1 + 2/3^2 + .3/3^3 + .. + 100/3^100
1/3*M= 1/3^2 + 2/3^3 + 3/3^4 + .. + 100/3^101
=> M- 1/3*C = 1/3^1 + (2/3^2 - 1/3^2) + (3/3^3 - 2/3^3) + .. + (100/3^100 - 99/3^100) - 100/3^101
=> 2/3*M = 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 - 100/3^101
+ xét S= 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 tương tự
1/3*S = 1/3^2 + 1/3^3 + 1/3^4 + .. + 1/3^101
=> S - 1/3*S = 1/3^1 - 1/3^101
<=> 2/3*S = (1/3 - 1/3^101)
<=> S = 3/2*(1/3 - 1/3^101) thay vào C ta có
2/3*M = 3/2*(1/3 - 1/3^101) - 100/3^101
<> M = 9/4*(1/3 - 1/3^101) - 150/3^101
<>M = 3/4 - 9/4*1/3^101 - 150/3^101 < 3/4
Thấy hay thì tíck cho mk 3 cái
\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3M-M=1+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)-\frac{100}{3^{100}}\)
\(2M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow M=1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow\frac{3}{2}< \frac{3}{4}\left(đpcm\right)\)