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Đặt \(A=5+5^2+5^3+....+5^{199}+5^{200}\)
\(\Leftrightarrow5A=5\left(5+5^2+5^3+....+5^{199}+5^{200}\right)\)
\(\Leftrightarrow5A=5^2+5^3+5^4+....+5^{200}+5^{201}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+5^4+....+5^{200}+5^{201}\right)-\left(5+5^2+5^3+....+5^{199}+5^{200}\right)\)
\(\Leftrightarrow4A=5^{201}-5\)
\(\Leftrightarrow A=\frac{5^{201}-5}{4}\)
\(B=\frac{2019}{1}+\frac{2018}{2}+\frac{2017}{3}+......+\frac{1}{2019}\)
\(=\left(\frac{2018}{2}+1\right)+\left(\frac{2017}{3}+1\right)+.....+\left(\frac{1}{2019}+1\right)+1\)
\(=\frac{2020}{2}+\frac{2020}{3}+\frac{2020}{4}+.....+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2020}\right)\)
\(=2020A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{2020A}=\frac{1}{2020}\)
a) Các số có dạng : \(\frac{1}{a\left(a+1\right)}=\frac{\left(a+1\right)-a}{a\left(a+1\right)}=\frac{1}{a}-\)\(\frac{1}{a+1}\)
Thế vào bởi các số sẽ có kết quả
b) Các số có dạng : \(\frac{1}{a\left(a+2\right)}=\frac{1}{2}.\frac{2}{a\left(a+2\right)}=\frac{1}{2}.\frac{\left(a+2\right)-a}{a\left(a+2\right)}\)\(=\frac{1}{2}.\left(\frac{1}{a}-\frac{1}{a+2}\right)\)
Làm tương tự trên
c) Lấy nhân tử chung là 5 rồi làm như câu a)
1x2x3x...2018x2019 - 1x2x3x..2018 - 1x2x3x4x...x2017x20182
= 1x2x3x...x2018x(2019 - 1 - 2018)
= 1x2x3x...x2018x0
= 0
a) Ta có: \(M=3+3^2+3^3+...+3^{2017}+3^{2018}+3^{2019}\)
\(=3.\left(1+3+3^2+3^3+...+3^{2016}+3^{2017}+3^{2018}\right)\)
\(\Rightarrow M⋮3\)
_Học tốt_
\(Q=\)\(1-2+3-4+...+2017-2018+2019\)
\(Q=\left(1-2\right)+\left(3-4\right)+...+\left(2017-2018\right)+2019\)
\(Q=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2019\)
\(Q=\left(-1\right).1009+2019\)
\(Q=\left(-1009\right)+2019\)
\(Q=1010\)
~~~~~~~~~~~~~~Hok tốt~~~~~~~~~~~~~~~~~
Đặt \(A=1-3^2+3^3-3^4+...+3^{2017}-3^{2018}+3^{2019}-3^{2020}\)
\(\Leftrightarrow A=1-\left(3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\right)\)
Đặt \(B=3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\)
\(3B=3\left(3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\right)\)
\(3B=3^3-3^4+3^5-....-3^{2018}+3^{2019}-3^{2020}+3^{2021}\)
\(3B+B=\left(3^3-3^4+3^5-....-3^{2018}+3^{2019}-3^{2020}+3^{2021}\right)\)
\(+\left(3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\right)\)
\(4B=3^{2021}+3^2\)
\(B=\frac{3^{2021}+3^2}{4}\)Thay vào A ta có A=\(1-\frac{3^{2021}+3^2}{4}\)
thank!