chung minh rang 1-1/2+1/3-1/4+.....+1/199-1/200=1/101+1/102+.....+1/200
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1-1/2+1/3-1/4+...+1/199-1/200=(1+1/2+1/3+1/4+...+199+1/200)-(1+1/2+1/3+...+1/100)=1+1/2+1/3+1/4+...+1/199+1/200-1-1/2-1/3-1/4-...-1/99-1/100=(1+1/2+1/3+...+1/100)-(1+1/2+1/3+...+1/100)+(1/101+1/102+...+1/200)=0+(1/101+1/102+...+1/200)=(1/101+1/102+...+1/200)(đpcm)
\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(VT=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(VT=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=VP\)=> ĐPCM
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(\text{đ}pcm\right)\)
Làm ơn giải giúp mình nhanh nhanh nhé, mình đang cần gấp, ai giải được mình k cho
bien doi so sdau tien
1-1/2+1/3-1/4+...+1/199-1/200
(1+1/3+1/5+...+1/199)-(1/2+1/4+1/6+...+1/200)
(1+1/2+1/3+1/4+1/5+1/6+...+1/199+1/200)-2*(1/2+1/4+1/6+,,,+1/200)
(1+1/2+1/3+1/4+...+1/200)-((1+1/2+1/3+...+1/100)
1/101+1/102+1/103+...+1/200=so bewn canh