cm 1/2+2/2^2+3/2^3+4/2^4+...+100/2^100<2
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Vì \(\dfrac{1}{a}\left(a>1\right)< 1với\forall a\)
mà \(2^2;3^2;.....;100^2>1\)
\(=>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow4A=2^2\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\)
\(\Rightarrow3A=4A-A=1+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^2}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)
\(\Rightarrow A=\left(1-\dfrac{1}{2^{100}}\right):3=\dfrac{1}{3}-\dfrac{1}{2^{100}.3}< \dfrac{1}{3}\left(đpcm\right)\)
Sửa đề: A=1/2^2+...+1/100^2
1/2^2<1/1*2
1/3^2<1/2*3
...
1/100^2<1/99*100
=>A<1-1/2+1/2-1/3+...+1/99-1/100
=>A<99/100<1
Đặt \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=2-\frac{100}{2^{99}}+\frac{100}{2^{100}}< 2-\frac{100}{2^{100}}+\frac{100}{2^{100}}=2\)
\(\Rightarrow A< 2\Leftrightarrow\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}< 2\left(đpcm\right).\)
cảm ơn nhé