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Gọi tử số của \(S\)là :\(A=1+2+2^2+2^3+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(2A-A=\left(2+2^2+2^3+...2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)
\(A=1-2^{2016}\)
\(\Rightarrow S=\frac{1-2^{2016}}{1-2^{2016}}=1\)
Ta có:
1/2^2+1/3^2+.....+1/20^2>1/2.2+1/3.4+1/4.5+.....+1/20.21
=1/4+1/3-1/21
=1/4+6/21
=45/84>1/2
Ta có:
1/2^2+1/3^2+..........+1/20^2<1/1.2+1/2.3+.....+1/19.20
=1-1/20
=19/20<1
ta có: C = 1/32 + 1/34 + 1/36 +...+ 1/3100 => 9C = 1 + 1/32 +1/34 +...+1/398
=> 9C - C = (1 + 1/32 + 1/34 +...+1/398 ) - (1/32 +1/34 + 1/36 +...+ 1/3100)
=> 8C = 1 - 1/3100 => C = (1 - 1/3100 ) / 8
đúng ko nhỉ
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^{3.}}+.............+\frac{1}{2^{100}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+.................+\frac{1}{2^{99}}\)
\(2B-B=1-\frac{1}{2^{100}}\)
\(B=1-\frac{1}{2^{100}}\)
\( C=\frac{1}{2}-\frac{1}{2^2}+.................+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(2 C=1-\frac{1}{2}+......................+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(2 C+C=1-\frac{1}{2^{100}}\)
\(C=\left(1-\frac{1}{2^{100}}\right):3\)
999 - 888 - 111 + 111 - 111 + 111 - 111
= 111 - 111 + 111 - 111 + 111 - 111
= 0 + 111 - 111 + 111 - 111
= 111 - 111 + 111 - 111
= 0 + 111 - 111
= 111 - 111
= 0
\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{2^2}=\frac{1}{2\times2}< \frac{1}{1\times2}\)
\(\frac{1}{3^2}=\frac{1}{3\times3}< \frac{1}{2\times3}\)
\(\frac{1}{4^2}=\frac{1}{4\times4}< \frac{1}{3\times4}\)
\(...\)
\(\frac{1}{99^2}=\frac{1}{99\times99}< \frac{1}{98\times99}\)
\(\frac{1}{100^2}=\frac{1}{100\times100}=\frac{1}{99\times100}\)
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
=>\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
=>\(A=2A-A=2+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
\(A=2+\frac{1}{2^{98}}\)
Vậy: \(A=2+\frac{1}{2^{98}}\)
Gọi \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2B-B=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow B=2-\frac{1}{2^{100}}\)
\(\Rightarrow A=2\)
Vậy A = 2