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a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)
Đặt \(B=2+2^2+...+2^{2017}\)
\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)
\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)
\(\Rightarrow B=2^{2018}-2\)
\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)
\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)
\(\Rightarrow A=-2\)
b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)
\(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)
\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)
\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)
\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)
Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)
\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)
\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)
\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)
\(\Rightarrow2016S< 2017\)
\(\Rightarrow S< \frac{2017}{2016}\)
\(\Rightarrow2016A< \frac{2017}{2016}\)
\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)
Ta có : S =\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\)\(-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)\(-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
\(\Rightarrow S=P\)
Khi đó : \(\left(S-P\right)^{2018}=0^{2018}=0\)
k chi mik nha!
-.-
S1 = 1-2+3-4+....+2017-2018
= (-1)+(-1)+....+(-1)
= (-1) x 1009
= -1009
Vì đề con viết thiếu nên cô đã sửa nhé.
Ta có \(S=1-2+2^2-2^3+...-2^{2017}\)
\(\Rightarrow4S=2^2.S=2^2\left(1-2+2^2-2^3+...-2^{2017}\right)\)
\(\Rightarrow4S=2^2-2^3+2^4-2^5+...-2^{2017}+2^{2018}-2^{2019}\)
\(\Rightarrow4S=S+1+2^{2018}-2^{2019}\)
\(\Rightarrow3S=1+2^{2018}-2^{2019}\)
\(\Rightarrow M=3S-2^{2018}=1-2^{2019}\)