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Đề \(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\) chứ bn!
Bài làm
ta có: 1.98 +2.97 + 3.96 + 98.1
= 1 + (1+2) + ( 1+2+3) +...+ ( 1+2+3+...+ 98)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+...+98.99}{2}\)
\(\Rightarrow A=\frac{1.98+2.99+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{\left(1.2+2.3+3.4+...+98.99\right).\frac{1}{2}}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{1}{2}\left(đpcm\right)\)
Lời giải:
Xét tử số:
$1+(1+2)+(1+2+3)+...+(1+2+3+....98)$
$=\underbrace{(1+1+.....+1)}_{98}+\underbrace{(2+2+...+2)}_{97}+....+\underbrace{97+97}_{2}+98$
$=1.98+2.97+3.96+...+98.1$
Do đó: $B=\frac{1.98+2.97+3.96+..+98.1}{1.98+2.97+3.96+...+98.1}=1$
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
\(\frac{\frac{1.2}{2}+\frac{2.3}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}=\frac{1}{2}\)
Ta gọi:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+97\cdot98}\) là A
\(=\frac{1+3+6+10+...+4753}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
\(\Rightarrow2A=\frac{2+6+12+20+...+9506}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
\(=\frac{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
=> 2A = 1
=> A = 1/2
Mik nghĩ bỏ 98.98 ở phần MS thì bài mới đúng, bn nên hỏi lại thầy bn
Vậy A = 1/2
