CMR: A = \(\frac{1}{2!}+\frac{2}{3!}+......+\frac{99}{100!}<1\)
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S
17 tháng 1 2018
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(3A+A=4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4A< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(B+3B=4B=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow B< \frac{3}{4}\) (2)
Từ (2) và (2) => \(4A< B< \frac{3}{4}\Rightarrow A< \frac{3}{16}\) (đpcm)
18 tháng 1 2018
\(A=\frac{7n-1}{4};B=\frac{5n+3}{12}\)
Tìm n để A,B đồng thời là các số nguyên tố
DD
19 tháng 3 2016
a)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...-\frac{1}{2^6}=A\)
2A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}\)
2A + A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^6}\)
3A = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)(đpcm)
AH
Akai Haruma
Giáo viên
4 tháng 10 2024
Lời giải:
$A=\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}-....+\frac{99}{2^{99}}-\frac{100}{2^{100}}$
$2A=1-\frac{2}{2}+\frac{3}{2^2}-....+\frac{99}{2^{98}}-\frac{100}{2^{99}}$
$\Rightarrow A+2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...-\frac{1}{2^{99}}-\frac{100}{2^{100}}$
$\Rightarrow 3A+\frac{100}{2^{100}}=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...-\frac{1}{2^{99}}$
$\Rightarrow 2(3A+\frac{100}{2^{100}}) =2-1+\frac{1}{2}-\frac{1}{2^2}+...-\frac{1}{2^{98}}$
$\Rightarrow 3A+\frac{100}{2^{100}}+2(3A+\frac{100}{2^{100}})=2-\frac{1}{2^{99}}$
$\Rightarrow 9A+\frac{300}{2^{100}}=2-\frac{1}{2^{99}}$
$\Rightarrow 9A=2-\frac{1}{2^{99}}-\frac{300}{2^{100}}<2$
$\Rightarrow A< \frac{2}{9}$
27 tháng 6 2020
Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+....\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....\frac{1}{3^{98}}-\frac{1}{3^{99}}\Rightarrow4A< B\left(1\right)\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+....\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\Rightarrow4B< 3\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4A< B< \frac{3}{4}\Rightarrow4A< \frac{3}{4}\Rightarrow A< \frac{3}{4}:4\Rightarrow A< \frac{3}{4}.\frac{1}{4}\Rightarrow A< \frac{3}{16}\)
=> đpcm.
\(\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{100-1}{100!}=\left(\frac{2}{2!}-\frac{1}{2!}\right)+\left(\frac{3}{3!}-\frac{1}{3!}\right)+....+\left(\frac{99}{100!}-\frac{1}{100!}\right)\)
\(=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+...+\left(\frac{1}{99!}-\frac{1}{100!}\right)=\left(\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{99!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)=\frac{1}{1!}-\frac{1}{100!}=\frac{100!-1}{100!}\)