so sánh: 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^99 +1/3^100 và 1/2
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VH
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so sánh S = 1/3 - 2/3^2 + 3/3^3 -4/3^4 + ... + 99/3^99 -100/3^100 và 1/5
NM
9 tháng 5 2022 lúc 20:42
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
H24
2 tháng 4 2023 lúc 19:23
so sánh
P=\(\dfrac{1+7^2+7^3+...+7^{100}}{1+7^2+7^3+...+7^{99}}\)
Q=\(\dfrac{1+9^2+9^3+...+9^{100}}{1+9^2+9^3+...+9^{99}}\)
So sánh:1/1×2+1/2×3+...+1/99×100 và 1
Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)\(< 1\)
Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}< 1\)
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Đặt :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy \(A=\frac{99}{100}\)
Vì \(\frac{99}{100}< 1\)nên \(A< 1\)
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\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
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so sánh a và b, biết
a = 100 - ( 1 + 1/2 = 1/3 + ... + 99/100) b = 1/2 + 2/3 + 3/4 +... +99/100
So sánh
a, 1/3 + 1/3^2 + 1/3^3 +....+ 1/3^99 + 1/3^100 và 1/2
b, 3/1^2*2^2 + 5/2^2 *3^2 +7/3^2*4^2 +......+ 19/9^2*10^2 và 1
a)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
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\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{99}}+\frac{1}{3^{100}}\) và \(\frac{1}{2}\)
So sánh
ta có: \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}+\frac{1}{3^{101}}\)
\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^{101}}< \frac{1}{3}\)
\(\Rightarrow\frac{2}{3}A< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)
\(\Rightarrow A< \frac{1}{2}\)
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A=1+1/2+2/3+3/4+......+99/100 so sánh với B=100-(1/2+1/3+1/4+......+1/100)
giúp mik với
ta có
\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=A\)
Vậy A=B
So sánh 1/31+1/32+1/33+...+1/399+1/3100 với 1/2
Giúp với cần gấp!!!!!!!
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3A=3\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A-A=2A\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^1}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(=1-\frac{1}{3^{100}}\)
\(2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3A=3\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A-A=2A\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^1}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(=1-\frac{1}{3^{100}}\)
\(2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)