A=1/2! +2/3! +3/4!+...+9/10!
Chứng minh A<1.
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a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+...+\frac{1}{5}\right)\)
\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{10}\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{5}\right)\)
Vậy A = B và A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
1/ A= \(\left(\frac{1}{1.2}\right)+\left(\frac{1}{3.4}\right)+...+\left(\frac{1}{9.10}\right)\)
B=(1/1+1/2+1/3+...+1/10)- (1/1+1/2+...+1/5)
<=> B=1/6+1/7+1/8+1/9+1/10.
A = 3 + 32 + 33 + ........... + 39 + 310
A = (3 + 32) + (33 + 34) + ................ + (39 + 310)
A = 3.(1 + 3) + 33.(1 + 3) + .................. + 39.(1 + 3)
A = 3.4 + 33.4 + ..................... + 39 . 4
A = 4.(3 + 33 + ................... + 39)
3A = 3^2+3^3+3^4+...+3^10+3^11
3A-A= [3^2-3^2]+[3^3-3^3]+...+[3^10-3^10]+3^11-3
2A=3^11-3
2A=177144
A=177144:2=88572
Vì dấu hiệu chia hết cho 4 là 2 chữ số tận cùng chia hết cho 4 thì số đó chia hết cho 4.
Ma : 72 chia het cho 4
Vay 88572 chia hết cho 4 hay A chia hết cho 4
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-1-\frac{1}{2}-...-\frac{1}{5}\)
\(=\frac{1}{6}+\frac{1}{7}+...+\frac{1}{10}\left(đpcm\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\)\(\frac{1}{10}\)
\(A=\frac{1}{1}+\frac{1}{3}+...+\frac{1}{9}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{10}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}-2.\frac{1}{2}-2.\frac{1}{4}-...-2.\frac{1}{10}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}-1-\frac{1}{2}-...-\frac{1}{5}\)
\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\left(đpcm\right)\)
~~~Hok tốt~~~
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{10}{10!}-\frac{1}{10!}\)
\(A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+...+\frac{1}{9!}-\frac{1}{10!}\)
\(A=1-\frac{1}{10!}\)
\(\Rightarrow A< 1\left(đpcm\right)\)