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\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)
\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)
\(B=\frac{2016}{-1}=-2016\)
P(x) = x2019 - 1000x2018 + 1000x2017 - 1000x2016 + ... + 1000x - 1
Với x = 999 => 1000 = x + 1
=> P(999) = x2019 - ( x + 1 )x2018 + ( x + 1 )x2017 - ( x + 1 )x2016 + ... + ( x + 1 )x - 1
= x2019 - x2019 - x2018 + x2018 + x2017 - x2017 - x2016 + ... + x2 + x - 1
= x - 1 = 999 - 1 = 998
Vậy ...
a) Đặt \(A=2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2-2^1\)
\(\Rightarrow2A=2^{2017}-2^{2016}+2^{2015}-2^{2014}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{2017}-2^{2015}+2^{2014}-2^{2013}+...+2^3-2^2\right)+\left(2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2+2^1\right)\)
\(\Rightarrow3A=2^{2017}+1\)
\(\Rightarrow A=\frac{2^{2017}+1}{3}\)
b) Đặt \(B=3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\)
\(\Rightarrow3B=3^{1001}-3^{1000}+3^{999}-3^{997}+...+3^3-3^2+3^1\)
\(\Rightarrow3B+B=\left(3^{1001}-3^{1000}+3^{999}-3^{998}+...+3^3-3^2+3^1\right)+\left(3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\right)\)
\(\Rightarrow4B=3^{1001}+3^0\)
\(\Rightarrow B=\frac{3^{1001}+1}{4}\)
a) Đặt A = 22016 - 22015 + 22014 - 22013 + ... + 22 - 21
2A = 22017 - 22016 + 22015 - 22014 + ... + 23 - 22
2A + A = (22017 - 22016 + 22015 - 22014 + ... + 23 - 22) + (22016 - 22015 + 22014 - 22013 + ... + 22 - 21)
3A = 22017 - 21
3A = 22017 - 2
\(A=\frac{2^{2017}-2}{3}\)
b) lm tương tự câu a
\(P=\frac{3^{2016}-6^{2016}+9^{2016}-12^{2016}+15^{2016}-18^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\frac{\left(1.3\right)^{2016}-\left(2.3\right)^{2016}+\left(3.3\right)^{2016}-\left(4.3\right)^{2016}+\left(5.3\right)^{2016}-\left(6.3\right)^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\frac{1^{2016}.3^{2016}-2^{2016}.3^{2016}+3^{2016}.3^{2016}-4^{2016}.3^{2016}+5^{2016}.3^{2016}-6^{2016}.3^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\frac{-3^{2016}\left(-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}\right)}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=-3^{2016}\)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)
Vậy B = - 2016
Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?