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mình gợi ý nè : bạn thử lấy T nhân với 2 xem ( cả hai vế nhé )
Nếu bạn không ra thì k cho mình đi mình trình bày cho đôn giản mà mỗi tội hơi dài một chút.
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
1) Đặt dãy trên là \(A\)
Theo bài ra ta có :
\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)
2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )
\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
1) Ta có B =
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(\frac{99}{100}\)
=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)
Sai thì thôi chứ mk chỉ làm rờ thôi
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
1/2T=2/22 +3/23 +4/24 +...+2017/22017
T-1/2T= (2/21+3/22+4/23+...+2017/22016)-(2/22+3/23+4/24+...+2017/22017)
1/2T=2/21+3/22+4/23+...+2017/22016-2/22-3/23-4/24-...-2017/22017
1/2T=1+(3/22-2/22)+(4/23-3/23)+...+(2017/22016-2016/22016)-2017/22017
1/2T=1+(1/22+1/23+1/24+...+1/22016)-2017/22017
xét A = 1/22+1/23+1/24+...+1/22016
phần này dễ bạn tự làm nhé
A=1/2-1/22016<1/2(vì 1/22016>0)
1/2T<1/21+1/2-(1/22016+2017/22017)
1/2T<3/2(vì 1/22016+2017/22017>0)
T<3/2:1/2
T<3
vậy T<3
1/2T=2/22 +3/23 +4/24 +...+2017/22017 T-1/2T= (2/21+3/22+4/23+...+2017/22016 )-(2/22+3/23+4/24+...+2017/22017 ) 1/2T=2/21+3/22+4/23+...+2017/22016 -2/22 -3/23-4/24 -...-2017/22017 1/2T=1+(3/22 -2/22 )+(4/23 -3/23 )+...+(2017/22016 -2016/22016 )-2017/22017 1/2T=1+(1/22+1/23+1/24+...+1/22016 )-2017/22017 xét A = 1/22+1/23+1/24+...+1/22016 phần này dễ bạn tự làm nhé A=1/2-1/22016<1/2(vì 1/22016>0) 1/2T<1/21+1/2-(1/22016+2017/22017 ) 1/2T<3/2(vì 1/22016+2017/22017>0) T<3/2:1/2 T<3