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= 1.2.3.....99/2.3.4....100
=1/100
k mk nha đáp án đúng đó
S = \(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
Đặt A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
2A = \(2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
= \(2+1+\frac{1}{2}+....+\frac{1}{2^8}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(A=2-\frac{1}{2^9}\)
\(\Rightarrow S=3\left(2-\frac{1}{2^9}\right)=\frac{3.\left(2^{10}-1\right)}{2^9}\)
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
Lời giải:
$A=\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}-....+\frac{99}{2^{99}}-\frac{100}{2^{100}}$
$2A=1-\frac{2}{2}+\frac{3}{2^2}-....+\frac{99}{2^{98}}-\frac{100}{2^{99}}$
$\Rightarrow A+2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...-\frac{1}{2^{99}}-\frac{100}{2^{100}}$
$\Rightarrow 3A+\frac{100}{2^{100}}=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...-\frac{1}{2^{99}}$
$\Rightarrow 2(3A+\frac{100}{2^{100}}) =2-1+\frac{1}{2}-\frac{1}{2^2}+...-\frac{1}{2^{98}}$
$\Rightarrow 3A+\frac{100}{2^{100}}+2(3A+\frac{100}{2^{100}})=2-\frac{1}{2^{99}}$
$\Rightarrow 9A+\frac{300}{2^{100}}=2-\frac{1}{2^{99}}$
$\Rightarrow 9A=2-\frac{1}{2^{99}}-\frac{300}{2^{100}}<2$
$\Rightarrow A< \frac{2}{9}$
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.......+\frac{1}{100^2}<\frac{1}{2}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)<\(\frac{1}{0.2}+\frac{1}{2.4}+\frac{1}{4.6}+.......+\frac{1}{98.100}\)
\(S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{49}{100}<\frac{1}{2}\)
Vậy \(\frac{49}{100}<\frac{1}{2}\)
Ta có 1/22<1/2*3
1/42<1/3*4
. . .
1/1002<1/99*100
=> S<1/2*3+1/3*4+...+1/99*100
=> S<1/2-1/3+1/3-1/4+...+1/99-1/100
=>S<1/2-1/100
=>S<49/100
Mà 49/100<1/2
=>S<1/2
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\)
ta có:
A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+....+\(\frac{1}{100^2}\)< B=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{99.100}\)
B=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+........+\(\frac{1}{99}\)-1/100
B=1-1/100=99/100<1
Vì a<b mà B lại bé hơn 1 =>A<1