Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+......................+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.....................+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...................+\dfrac{101}{3^{101}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..............+\dfrac{100}{3^{99}}\right)\)\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+................+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+.............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)-\left(1+\dfrac{1}{3}+..........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Chúc bn học tốt ~
1/2^2=4
1/3^2<1/2.3
.................
1/100^2<1/99.100
A<1/4+1/2.3+...+1/99.100
A<1/4+1/2-1/100
A<1/4<3/4
Vậy A<3/4(dpcm).CHÚC BẠN HỌC TỐT!
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
Kiyoko Vũ
a, xét từng đoạn 1 , 1/2 ,1/2^3 ,1/2^4 ,1/2^5 ,1/2^6
ta có
1 = 1
1/2 + 1/3 < 1/2 + 1/2 = 1
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1
tương tự
1/16 +1/17 + .. + 1/31 < 1
1/32 + 1/33 + .. + 1/63 < 1
=> cộng lại => A < 6
b, Câu hỏi của trịnh quỳnh trang - Toán lớp 6 - Học toán với OnlineMath
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
Chứng minh rằng:
\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\)<1
Ta có: 1/22 < 1/ 1.2
1/32 < 1/2.3
1/42 < 1/3.4
....
1/ 1002 < 1/ 99.100
Nên A< 1/1.2+1/2.3+...+1/99.100
= 1- 1/2+1/2 -1/3+1/3 -1/4+...+1/99-1/100
= 1- 1/100
<1 Vậy A><1. >
Ma 1 > 1/100
Vay…
\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}221+321+421+...+10021<1.21+2.31+3.41+...+99.1001
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1=1−21+21−31+31−41+...+991−1001=1−1001<1
D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
=>3D=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>3D-D=(1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\))
=>2D=1-\(\dfrac{1}{3^{100}}< 1\)
=>D<\(\dfrac{1}{2}\)
Vậy...