Cho A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\).So sánh A với 1
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DV
10
TM
11 tháng 5 2016
Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2016}}\right)\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{2016}}\)
\(\Rightarrow A<1\left(đpcm\right)\)
HP
11 tháng 5 2016
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2016}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2015}}\)
=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2016}}\right)\)
=>\(A=1-\frac{1}{2^{2016}}\)
Vậy \(A=1-\frac{1}{2^{2016}}\)
IM
16 tháng 8 2016
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right)..................\left(1^2-2016^6\right)}{2^2.3^2.4^2...........2016^2}\)
\(\Leftrightarrow A=\frac{\left(1-2\right)\left(1+2\right)\left(1-3\right)\left(1+3\right)........\left(1-2016\right)\left(1+2016\right)}{2^2.3^2..........2016^2}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(3\right)\left(-2\right)\left(4\right).............\left(-2015\right)\left(1017\right)}{\left(2.3.4......2016\right)\left(2.3.4.2016\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)......\left(-2015\right)\right]\left(3.4.....2017\right)}{\left(2.3.4....2016\right)\left(2.3.4...2017\right)}\)
\(\Leftrightarrow A=-\frac{1}{2016.2}=-\frac{1}{4032}>-\frac{2}{2016}\)
\(\Leftrightarrow A=-\frac{2}{2016}\)
IM
16 tháng 8 2016
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right)..........\left(1^2-2016^2\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(3\right)\left(-2\right)\left(4\right)....\left(-2015\right)\left(2017\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right).....\left(-2015\right)\right]\left(3.4.5...2017\right)}{\left(2.3.....2016\right)\left(2.3.4....2016\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)2017}{2016}=-\frac{2017}{2016}< \frac{1}{2}\)
=> A<1/2
23 tháng 4 2018
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
23 tháng 4 2018
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
7 tháng 9 2018
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
DA
24 tháng 6 2020
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\(=\frac{1.51}{50.2}=\frac{51}{100}\)
NC
1
NN
0
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{2016}}<1\)
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