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đặt A=32/20x23+32/23x26+...+32/77x80
\(A=3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3\cdot\frac{3}{80}\)
\(=\frac{9}{80}\)
A=1/20*23+1/23*26+...+1/77*80
=1/3(1/20-1/23+1/23-1/26+...+1/77-1/80)
=1/3*3/80=1/80<1/79
đặt A=9/20x23+9/23x26+...+9/77x80
<=>\(A=3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)
\(\Rightarrow A=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(\Rightarrow A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(\Rightarrow A=3\cdot\frac{3}{80}\)
\(\Rightarrow A=\frac{9}{80}\)
a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2007x2009}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2009}\right)=\frac{1}{2}\cdot\frac{2008}{2009}=\frac{1004}{2009}\)
....
các bài cn lại bn lm tương tự nha
b, \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
3A = \(\dfrac{1}{6}+\dfrac{1}{18}+...+\dfrac{1}{330}\)
3A-A = \(\dfrac{1}{6}-\dfrac{1}{990}\)
2A = 82/495
A =82/495 : 2
A=41/495
\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}\)
\(=\frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right]\)
\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+...+\frac{1}{77}-\frac{1}{80}\right]\)
\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)
\(=\frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)
\(=\frac{1}{3}\cdot\frac{3}{80}=\frac{1}{1}\cdot\frac{1}{80}=\frac{1}{80}\)
Mà \(\frac{1}{80}< \frac{1}{9}\)nên \(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}< \frac{1}{9}\)
Vậy : ...
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
Trong 3 số `2n+1, 2n+2, 2n+3` luôn có một số chia hết cho 3
\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮3\) (1)
Xét \(n⋮2\)
Có: \(2n⋮2,2⋮2\Rightarrow2n+2⋮2\)
\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮2\) (2)
Xét \(n⋮̸2\)
Có: \(2n⋮2\left(dư1\right),1⋮2\left(dư1\right)\Rightarrow2n+1⋮2\)
\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮2\) (3)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrowđpcm\)
\(S=1+2+2^2+...+2^{99}\)
\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)
\(S=3+2^2.3+...+2^{98}.3\)
\(=3\left(1+2^2+...+2^{98}\right)⋮3\)
Đặt \(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\) ta có :
\(A=3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3.\frac{3}{80}\)
\(A=\frac{9}{80}< 1\) ( tử bé hơn mẫu )
Vậy \(A< 1\)
Chúc bạn học tốt ~