Cho S = \(\frac{1}{^{2^2}}+\frac{1}{^{3^2}}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
Chứng minh: S <\(\frac{3}{4}\)
Bài 1: Chứng minh rằng: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 2: Cho \(n\in N;n>1\). Chứng minh rằng: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}\notin N\)
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Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
Cho \(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+..+n}\)
Chứng minh S<2
\(\frac{1}{1+2+3+...+n}=\frac{1}{\frac{\left(1+n\right).n}{2}}=\frac{2}{\left(1+n\right).n}=2.\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
áp dụng vào mà làm
Ta có công thức: \(1+2+3+....+n=\frac{n.\left(n+1\right)}{2}\)
Áp dụng vào tình tổng S:
\(S=1+\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+.....+\frac{1}{\frac{n.\left(n+1\right)}{2}}\)
\(S=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.....+\frac{1}{\frac{n.\left(n+1\right)}{2}}\)
\(S=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{n\left(n+1\right)}\)
Đặt \(A=\frac{2}{2.3}+\frac{2}{3.4}+.....+\frac{2}{n\left(n+1\right)}\) ,ta có:
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{2}-\frac{1}{n+1}=\frac{n+1-2}{2\left(n+1\right)}=\frac{n-1}{2n+2}\)
=>\(A=\frac{n-1}{2n+2}.2=\frac{2\left(n-1\right)}{2n+2}=\frac{2n-2}{2n+2}=\frac{2n+2-4}{2n+2}=1-\frac{4}{2n+2}<1\)
=>A < 1
Mà S=1+A
=>S < 2 (đpcm)
cho S=\(\frac{1}{4^1}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)
chứng minh S<\(\frac{1}{2}\)
^ là dấu phân số nhé
cho A=1^1.2+1^2.3+...+1^2014.2015
1^1.2>1^4; 1^2.3>2^42; 1^3.4>3^43;...;1^2014.2015>2014^42014
mà A=1^1.2+1^2.3+...+1^2104.2015=1-1^2+1^2-1^3+1^3+...+1^2014-1^2015
A=1-1^2015=2014^2015
mà 2014^2015>1^2>S nên 1^2>S
dạng 1 : so sánh
a) P = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}\)và Q = \(1\frac{3}{4}\)
dạng 2 : toán chứng minh
1. cho S = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\)chứng minh rằng : \(\frac{1}{4}< S< \frac{91}{330}\)
2. cho S = \(\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\). CMR : 3 < S < 8
3. CMR : \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
2.a) Vào question 126036
b) Vào question 68660
Cho S=\(\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{1992}{2^{1991}}\). Chứng minh rằng S<4
S = \(\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{1992}{2^{1991}}\)
2.S = \(2+\frac{2}{2^0}+\frac{3}{2^1}+...+\frac{1992}{2^{1990}}\)
=> 2.S - S = \(2+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}-\frac{1992}{2^{1991}}\)
=> S = \(2-\frac{1992}{2^{1991}}+\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}\right)\)
Đặt A = \(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}\)
=>2.A = 2 + \(\frac{1}{2^0}+\frac{1}{2^1}+...+\frac{1}{2^{1989}}\)
=> 2.A - A = 2 - \(\frac{1}{2^{1990}}\)=A
Vậy S = \(4-\frac{1}{2^{1990}}-\frac{1992}{2^{1991}}<4\)
tic cho tuiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Chứng minh rằng : S > 1
sửa đề : S < 1
\(s< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+..................+\frac{1}{9.10}\)
\(\Leftrightarrow S< 1-\frac{1}{10}\)
vậy S < 1
Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{9^2}\)
Chứng minh rằng: \(\frac{2}{5}< S< \frac{8}{9}\)
Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}.\)
Mà\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow S< \frac{8}{9}\)
Và \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
Mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow S>\frac{2}{5}\)
Vậy: \(\frac{2}{5}< S< \frac{8}{9}\)
Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
50 mũ 2 nhé
Chứng minh rằng S<\(\frac{3}{4}\)
\(S=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{50^2}\right)\)
Xét \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< \frac{1}{2}-\frac{1}{50}< \frac{1}{2}\)
\(=>A< \frac{1}{2}\)
=>\(S=\frac{1}{4}+A< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
vậy S<3/4