\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{127}\). Chứng minh A>4
a , cho A = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\) . Chứng minh A < \(\dfrac{7}{4}\)
b ,cho B = 21 + 22 + 23 + ... + 260 . Chứng minh B \(⋮\) 21
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
Cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+....+\(\dfrac{1}{100^2}\). Chứng minh rằng A<\(\dfrac{3}{4}\)
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:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
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 $A< \frac{3}{4}$
Chứng minh rằng:
a) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{2}{63}>2\)
b) \(\dfrac{3}{9\cdot14}+\dfrac{3}{14\cdot19}+\dfrac{3}{19\cdot24}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}< \dfrac{1}{15}\)
c) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
Cho A= \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+\dfrac{1}{10^2}+...+\dfrac{1}{160^2}\)
Chứng minh: \(\dfrac{1}{8}< A< \dfrac{3}{16}\)
cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\) chứng minh \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
=>\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
=>\(A>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
=>\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=>\(A>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5}{10}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
Do đó: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
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
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)
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)