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\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(B=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(B=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}=A\)
=>A/B=1
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Đầu tiên ta phân tích A
A = 1/1-1/2+1/3-1/4+...+1/99-1/100
sau đó chia vế A thành 2 phần
A = (1/1+1/3+...+1/99) - (1/2+1/4+...+1/100)
gọi (1/1+1/3+...+1/99) = a
gọi (1/2+1/4+...+1/100) = b
áp dụng tính chất (a-b) = (a+b) - 2b
=> A = (1/1+1+2+1/3+1/4+...+1/99+1/100) - 2(1/2+1/4+...+1/100)
=> A = (1/1+1+2+1/3+1/4+...+1/99+1/100) - (1/1+1/2+...+1/50)
=> A = 1/1-1/1+1/2-1/2+...+1/50-1/50+1/51+1/52+...+1/100
=> A = 1/51+1/52+...+1/100
vậy A / B = \(\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2011}{51}+\frac{2011}{52}+...+\frac{2011}{100}}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{2011\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}=2011\)
mà 2011 là số nguyên => (dpcm)
>>Dat Doan hơi nhầm nè, bạn phải ghi B/A chứ ko phải A/B; thành ra mới bằng 2011 chứ nếu A/B=1/2011 đó!!!
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Answer:
Mình làm thành tính tỉ số luôn nhé!
\(A=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}}\)
Ta xét \(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{100}-1-\frac{1}{2}-...-\frac{1}{50}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{50}-\frac{1}{50}\right)+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}\)
\(\Rightarrow\frac{A}{B}=1\)
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Xét mẫu số: 1/(2x3) + 1/(3x4) + …… + 1/(99x100)
= 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100
= (1 + 1/3 + ............ + 1/99) – (1/2 + 1/4 + .......... + 1/100)
= (1 + 1/3 + ............ + 1/99)+(1/2+1/4+1/6+….+1/100) – (1/2+1/4+1/6+ .......... + 1/100)x2
= (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2 + 1/3 + ....... +1/50 )
= 1/51 + 1/52 + 1/53 + ............. + 1/100 (Đơn giản số trừ)
Vậy: (1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/1x2 + 1/3x4 + .......... + 1/99x100) =
(1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/51 + 1/52 + 1/53 + ............. + 1/100) = 1