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Ta có:
B = (1/2^2 - 1) . (1/3^2 - 1) ... (1/100^2 - 1)
B = (1/4 - 1) . (1/9 - 1) ... (1/10000 - 1)
B = -3/4 . (-8/9) ... (-9999/10000)
B = -(3/4 . 8/9 ... 9999/10000) (Vì có 99 thừa số âm nên tích trên có giá trị âm)
B = -(1.3/2.2 . 2.4/3.3 ... 99.101/100.100)
B = -(1.2.3.4...99/2.3.4.5...100 . 3.4.5...101/2.3.4.5...100)
B = -(1/100 . 101/2)
B = -101/200
Mà -1/2 = -100/200
Vì -101 < -100 nên -101/200 < -100/200
Vậy B < -1/2
P/S: Mình nghĩ là vậy nhưng không chắc là đúng không nữa. Nếu sai thì cho mình xin lỗi nha! ^_^
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
\(B=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(B=-\frac{3}{2^2}.\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right)...\left(-\frac{9999}{100^2}\right)\)
\(B=-\left(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\right)\)(Vì có 99 thừa số, mỗi thừa số là âm nên kết quả là âm)
\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\right)\)
\(B=-\left(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\right)\)
\(B=-\left(\frac{1}{100}.\frac{101}{2}\right)\)
\(B=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
\(\Rightarrow B< -\frac{1}{2}\)
Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
Ta có : \(\frac{1}{n^2}-1=\frac{1-n^2}{n^2}=\frac{\left(1-n\right)\left(1+1\right)}{n^2}\)
Áp dụng :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)
\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)
Sr còn thiếu
\(A=-\frac{2015}{4028}< \frac{-2014}{4028}=-\frac{1}{2}\)
Vậy \(A< B\)
Đặt \(100=n\) , ta có :
\(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{n^2}-1\right)\)
\(=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.....\frac{\left(1-n\right)\left(1+n\right)}{n^2}\)
\(=\frac{\left(-1\right).\left(-2\right)....\left(1-n\right)}{2.3.....n}.\frac{3.4........\left(1+n\right)}{2.3.....n}\)
\(=\frac{\left(-1\right).2.3.....\left(n-1\right)}{2.3......n}.\frac{3.4.....\left(n+1\right)}{2.3.......n}\)
\(=\frac{\left(-1\right)}{n}.\frac{n+1}{2}=\frac{-1}{2}.\frac{n+1}{n}< \frac{-1}{2}\)
Vậy \(B< \frac{-1}{2}\)