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Số số hạng của vế trái là :
(a - 1) : 1 + 1 = a (số hạng)
Suy ra : \(\dfrac{\left(a+1\right)a}{2}=4095\), từ đó :
\(\left(a+1\right)a=4095\) x \(2=8190\)
Ta có : 90 x 91 = 8190 nên a = 90
Đ/s : 90
=>1-1/2+1/2-1/3+...+1/a-1/(a+1)=2020/2021
=>1-1/(a+1)=2020/2021
=>1/(a+1)=1/2021
=>a+1=2021
=>a=2020
b)ta đặt A: \(A=\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\)
\(A=\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+..+\left(\frac{98}{2}+1\right)+\left(\frac{99}{1}-98\right)\)
\(A=\frac{100}{99}+\frac{100}{98}+..+\frac{100}{2}+\frac{100}{100}\)
\(A=100\cdot\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+..+\frac{1}{2}\right)\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)
\(3A=1-\frac{1}{64}\)
\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)
1/2:0,5-1/4:0,25+1/8:0,125-1/10:0,1
=0,5:0,5-0,25:0,25+0,125:0,125-0,1:0,1
=1-1+1-1
=0
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2021}\)
\(A=\frac{1}{\frac{2\times3}{2}}+\frac{1}{\frac{3\times4}{2}}+\frac{1}{\frac{4\times5}{2}}+...+\frac{1}{\frac{2021\times2022}{2}}\)
\(A=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2021\times2022}\right)\)
\(A=2\times\left(\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+...+\frac{2022-2021}{2021\times2022}\right)\)
\(A=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2022}\right)\)
\(A=2\times\left(\frac{1}{2}-\frac{1}{2022}\right)\)
\(A=\frac{1010}{1011}\)