Chứng minh: \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^n}< 1\)
Những câu hỏi liên quan
Cho N = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\).Chứng minh N <1
Chứng minh: Adfrac{2^3+1}{2^3-1}.dfrac{3^3+1}{3^3-1}.dfrac{4^3+1}{4^3-1}....dfrac{9^3+1}{9^3-1} dfrac{3}{2}
Bdfrac{1}{2!}+dfrac{1}{3!}+dfrac{1}{4!}+....+dfrac{1}{n!} 1
Cdfrac{1}{2!}+dfrac{2}{3!}+dfrac{3}{4!}+....+dfrac{n-1}{n!} 1
Dleft(1-dfrac{2}{6}right)left(1-dfrac{2}{12}right)left(1-dfrac{2}{20}right)....left(1-dfrac{2}{nleft(n+1right)}right)dfrac{1}{3}
Đọc tiếp
Chứng minh: \(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}.\dfrac{4^3+1}{4^3-1}....\dfrac{9^3+1}{9^3-1}< \dfrac{3}{2}\)
\(B=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{n!}< 1\)
\(C=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+....+\dfrac{n-1}{n!}< 1\)
D=\(\left(1-\dfrac{2}{6}\right)\left(1-\dfrac{2}{12}\right)\left(1-\dfrac{2}{20}\right)....\left(1-\dfrac{2}{n\left(n+1\right)}\right)>\dfrac{1}{3}\)
1) Chứng minh rằng: 1+dfrac{1}{2sqrt{2}}+dfrac{1}{3sqrt{3}}+...+dfrac{1}{nsqrt{n}} 2sqrt{2}left(nin Nright)
2) Chứng minh rằng: dfrac{2}{3}+sqrt{n+1} 1+sqrt{2}+sqrt{3}+...+sqrt{n} dfrac{2}{3}left(n+1right)sqrt{n}
3) 2sqrt{n}-3 dfrac{1}{sqrt{2}}+dfrac{1}{sqrt{3}}+...+dfrac{1}{sqrt{n}} 2sqrt{n}-2
4) dfrac{sqrt{2}-sqrt{1}}{2+1}+dfrac{sqrt{3}-sqrt{2}}{3+2}+...+dfrac{sqrt{n+1}-sqrt{n}}{n+1+n} dfrac{1}{2}left(1-dfrac{1}{sqrt{n+1}}right)
Đọc tiếp
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Chứng minh:
Adfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+...+dfrac{1}{18.19.20} dfrac{1}{4}
Bdfrac{36}{1.3.5}+dfrac{36}{5.7.9}+dfrac{36}{9.11.13}+...+dfrac{36}{25.27.29} 3
Cdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{n^2}in 1left(nin N,nge2right)
Ddfrac{1}{4^2}+dfrac{1}{6^2}+dfrac{1}{8^2}+...+dfrac{1}{left(2nright)^2} 4left(nin N,nge2right)
Edfrac{2!}{3!}+dfrac{2!}{4!}+dfrac{2!}{5!}+...+dfrac{2!}{n!} 1left(nin N,nge3right)
Đọc tiếp
Chứng minh:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{5.7.9}+\dfrac{36}{9.11.13}+...+\dfrac{36}{25.27.29}< 3\)
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\in< 1\left(n\in N,n\ge2\right)\)
\(D=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< 4\left(n\in N,n\ge2\right)\)
\(E=\dfrac{2!}{3!}+\dfrac{2!}{4!}+\dfrac{2!}{5!}+...+\dfrac{2!}{n!}< 1\left(n\in N,n\ge3\right)\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
Đúng 0
Bình luận (1)
Chứng minh\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)
Chứng minh \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004 }}< 0.2\)
a)
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..........................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)
\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)
\(\Leftrightarrow S< 0,2\left(đpcm\right)\)
Đúng 0
Bình luận (0)
2. Chứng minh
a, dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{50^2} 1
b, dfrac{1}{3} dfrac{1}{101}+dfrac{1}{102}+dfrac{1}{103}+...+dfrac{1}{150} dfrac{1}{2}
Đọc tiếp
2. Chứng minh
a, \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{50^2}\) < 1
b, \(\dfrac{1}{3}\)< \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{150}\)< \(\dfrac{1}{2}\)
Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
Đúng 1
Bình luận (0)
2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
Đúng 1
Bình luận (0)
2b.
Gọi tổng trên là $T$
Chứng minh vế đầu tiên:
Ta có:
$\frac{1}{101}> \frac{1}{150}$
$\frac{1}{102}> \frac{1}{150}$
....
$\frac{1}{149}> \frac{1}{150}$
$\Rightarrow T> \underbrace{\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}}_{50}=\frac{50}{150}=\frac{1}{3}$ (đpcm)
Chứng minh vế số 2:
$\frac{1}{101}< \frac{1}{100}$
$\frac{1}{102}< \frac{1}{100}$
....
$\frac{1}{150}< \frac{1}{100}$
$\Rightarrow T< \underbrace{\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}}_{50}=\frac{50}{100}=\frac{1}{2}$ (đpcm)
Đúng 0
Bình luận (0)
Chứng Minh Rằng : A= \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\) <\(\dfrac{3}{4}\)
Lời giải:
$A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}$
$A< \frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{1}{4}+\frac{1}{2}$
Hay $A< \frac{3}{4}$
Đúng 0
Bình luận (0)
Chứng minh rằng
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\)
\(Vì\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};.....;\dfrac{1}{n^2}< \dfrac{1}{(n-1).n}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{\left(n-1\right).1}< 1\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\left(đpcm\right)\)
vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\)
Đúng 0
Bình luận (1)
ĐặtA= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}\)
Do \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.............
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta suy ra : A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{\left(n-1\right)n}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-....-\dfrac{1}{\left(n-1\right)}+\dfrac{1}{n-1}-\dfrac{1}{n}\)
=\(1-\dfrac{1}{n}\)
Mà 1-\(\dfrac{1}{n}\)<1
=> A<1 (đpcm)
Đúng 0
Bình luận (0)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
mà \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=1-\dfrac{1}{n}< 1\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) (đpcm)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
1/ Cho A dfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+.....+dfrac{99}{3^{99}}-dfrac{100}{3^{100}} Chứng minh A dfrac{3}{16}2/ Cho B(dfrac{1}{2^2}-1)(dfrac{1}{3^2}-1)....(dfrac{1}{100^2}-1) So sánh B và dfrac{-1}{2}
Đọc tiếp
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
Đúng 1
Bình luận (0)