cho s = 1/5^2-1/5^3+1/5^4-1/5^5+...-15^101.chung minh s<1/30.
giai giup minh voi
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B = \(1+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{630}=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{1260}\)
B = \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{35.36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{35}-\frac{1}{36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{36}\right)=1+2.\frac{17}{36}\)
B = \(1+\frac{17}{18}\)
B = \(\frac{35}{18}\)
\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{99x101}\)
\(A\)\(x2=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}\)
\(A\)\(x2=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(A\)\(x2=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}:2=\frac{100}{101}x\frac{1}{2}=\frac{50}{101}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
...
\(\frac{1}{100^2}<\frac{1}{99.100}\)
===>\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{1}{2}\)
S = (1 / 31 + ... + 1 / 40) + (1 / 41 + ... + 1/ 50) + (1 / 51 + ... + 1 / 60) <
10 / 31 + 10 / 41 + 10 / 51 < 10 / 30 + 10 / 40 + 10 / 50 = 1 / 3 + 1 / 4 + 1 / 5 =
7 / 12 + 1 / 5 < 3 / 5 + 1 / 5 = 4 / 5
Tương tự:
S > 10 / 40 + 10 / 50 + 10 / 60 = 1 / 4 + 1 / 5 + 1 / 6 = 5 / 12 + 1 / 5 > 2 / 5 + 1 / 5 = 3 / 5
=> 3 / 5 < S < 4 / 5
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
S = 1/5^2 - 1/5^3 +1/5^4 -1/5^5 +... - 1/5^101
5^2.S = 5^2. ( 1/5^2 - 1/5^3 +1/5^4 -1/5^5 +... - 1/5^101)
25 .S =1-1/5^2+1/5^3-1/5^4+..-1/5^100
25S+S = (1-1/5^2+1/5^3-1/5^4+..-1/5^100)+( 1/5^2 - 1/5^3 +1/5^4 -1/5^5 +... - 1/5^101)
26S=1-1/5^101
Bạn tự làm tiếp
`Answer:`
\(S=\frac{1}{5^2}-\frac{1}{5^3}+\frac{1}{5^4}-\frac{1}{5^5}+...-\frac{1}{5^{101}}\)
\(\Rightarrow5S=\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...-\frac{1}{5^{101}}\)
\(\Rightarrow5S+S=\left(\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...-\frac{1}{5^{101}}\right)+\left(\frac{1}{5^2}-\frac{1}{5^3}+\frac{1}{5^4}-\frac{1}{5^5}+...-\frac{1}{5^{101}}\right)\)
\(\Rightarrow6S=\frac{1}{5}-\frac{1}{5^{101}}\)\(< \frac{1}{5}\)
\(\Rightarrow S< \frac{1}{30}\)