rút gọn biểu thức sau:
M=3-3^2+3^3-3^4+...+3^2015-3^2016
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\(M=3-3^2+3^3-3^4+...+3^{2015}-3^{2016}\)
\(3M=3^2-3^3+3^4-3^5+....+3^{2016}-3^{2017}\)
\(3M+M=3-3^{2017}\)
\(4M=3-3^{2017}\)
\(M=\frac{3-3^{2017}}{4}\)
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
Với mọi n>0 ta có:\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng đẳng thức trên vào D ta được:
\(D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}=1-\frac{\sqrt{2016}}{2016}=\frac{2016-\sqrt{2016}}{2016}\)
M = 3 - 32 +33 - 34+...+32015 - 32016
=> 3M = 32 - 33 + 34 - 35 +...+ 32016 - 32017
=> M + 3M = 3 - 32017
\(M=\frac{3-3^{2017}}{4}\)