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Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
Nên từ đây => \(A< 1\) (ĐPCM)
A=1/3+2/3^2+3/3^3+4/3^4+...+101/3^101
3A=1+2/3^1+3/3^2+4/3^3+...+101/3^100
3A-A=(1+2/3^1+3/3^2+4/3^3+...+101/3^100)-(1/3+2/3^2+3/3^3+4/3^4+...+101/3^101)
2A=(1-101/3^101)+(2/3-1/3)+(3/3^2-2/3^2)+...+(101/3^100-99/3^100)
2A=1-101/3^101+(1/3+1/3^2+,,,+1/3^100)
(1/3+1/3^2+,,,+1/3^100) đặt B
B=1/3+1/3^2+,,,+1/3^100
3B=1+1/3+...+1/3^99
3B-B=(1+1/3+...+1/3^99)-(1/3+1/3^2+,,,+1/3^100)
2B=1-1/3^100
B=(1-1/3^100):2=1/2-1/3^100.2
thay B vào 2A ta đc
2A=1-101/3^101+B
2A=1-101/3^101+1/2-1/3^100.2
2A=(1+1/2)-(101/3^101+1/3^100.2)
2A=3/2-(101/3^101+1/3^100.2)
A=3/4-(101/3^101+1/3^100.2)<3/4
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\left(\frac{100}{2}+1\right)+\left(\frac{99}{3}+1\right)+...+\left(\frac{1}{101}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\frac{102}{2}+\frac{102}{3}+...+\frac{102}{101}+\frac{102}{102}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{102.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}\right)}\)
\(A=\frac{1}{102}\)
Lời giải:
Ta có:
\(\text{VT}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\text{VP}\)
Ta có đpcm.
