So sánh A với 2 biết A = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\).
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DT
Đỗ Thanh Hải
CTVVIP
25 tháng 5 2021
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)
ĐPcm
25 tháng 5 2021
Giải:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1.
T
17 tháng 6 2019
#)Giải :
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)
\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)
\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)
Có lẽ đúng .........................
2 tháng 11 2016
Ta có
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)
Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)
\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)
\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)
Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)
Vậy A < 2
21 tháng 2 2018
Mình không chắc đã đúng đâu nhưng mình cứ giair thử nhé !
Ta có :
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)+ ... + \(\frac{1}{99}-\frac{1}{100}\)
= \(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...\frac{1}{99}\right)\)- \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{100}\right)\)
= \(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...\frac{1}{99}\right)\)+ \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{100}\right)\)
- \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)x 2
= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)- \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
= \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)= B
Vậy , A = B
~ Chúc bạn học giỏi ! ~
19 tháng 4 2017
A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
A < 1 - \(\frac{1.}{100}\)
A < \(\frac{99}{100}< \frac{199}{100}\)
=> A < \(\frac{199}{100}\)
b,
S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)
S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)
S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)
S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)
S = \(\frac{1.11}{2.10}\)
S = \(\frac{11}{20}\)
D
16 tháng 4 2017
Ta có: \(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{100}{2^{101}}\)
\(A-\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)
\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)
Vậy A<2
20 tháng 3 2017
2 vế bằng nhau
100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
100- 1-1/2-1/3-...-1/100 = 1/2+2/3+3/4+...+99/100
100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)
100 = 1 + 1 + 1 + ... + 1 (100 số hạng)
100 = 100
Vậy 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)