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a,1+(-2)+3+(-4)+..........+19+(-20)
=[1+(-2)]+[3+(-4)]+........+[19+(-20)]
=(-1)+(-1)+.......+(-1){Có 10 số (-1)}
=(-1).10
=-10
b,1-2+3-4+........+99-100
=(1-2)+(3-4)+........+(99-100)
=(-1)+(-1)+.........+(-1){ Có 50 số (-1)}
=(-1).50
=-50
Sai đề câu c,-1+3-5+7-9+.........+99-101
=(-1)+(3-5)+(7-9)+........+(99-101)
=(-1)+(-2)+(-2)+...........+(-2){Có 25 số (-2)}
=(-1)+(-2).25
=(-1)+(-50)
=-51
Hình như sai đề d,1+2-3-4+............+98-99-100+101
=1+(2-3-4+5)+........+(98-99-100+101)
=1+0+.........+0
=1
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Trả lời:
\(A=-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-\frac{1}{5^2}-...-\frac{1}{99^2}-\frac{1}{100^2}\)
\(=-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\right)\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
........
\(\frac{1}{99^2}< \frac{1}{98.99}\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< 1-\frac{1}{100}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\right)>-1\)
Vậy A > - 1
\(A=-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Ta có \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}< 1\)
=> A > -1
ta có
\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=A\)
Vậy A=B