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Đề đúng là \(3^{n+1}+2^{n+1}+3^{n-1}+2^{n-1}\)
\(=\left(3^{n+1}+3^{n-1}\right)+\left(2^{n+1}+2^{n-1}\right)\)
\(=3^{n-1}\left(3^2+1\right)+2^{n-2}\left(2^3+2\right)\)
\(=3^{n-1}.10+2^{n-2}.10\)
\(=10\left(3^{n-1}+2^{n-2}\right)\)chia hết cho 10
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.......+\left(\frac{1}{2}\right)^{2017}\)
\(\Rightarrow2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+......+\left(\frac{1}{2}\right)^{2016}\)
\(\Rightarrow2S-S=1-\left(\frac{1}{2}\right)^{2017}\)
\(\Rightarrow S=1-\left(\frac{1}{2}\right)^{2017}< 1\left(đpcm\right)\)
Vì \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}=1-\frac{1}{n+1}< 1\)=> Q < 1 (đpcm)
Vô lí vì C=1/3+1/3^2 +... luôn lớn hơn 1/3. Chắc là c/m <1/2 đúng ko
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
\(3C-C=1-\frac{1}{3^{2018}}\)
\(2C=1-\frac{1}{3^{2018}}\)
\(C=\frac{1}{2}-\frac{2}{3^{2018}}< \frac{1}{2}\)
Vậy \(C< \frac{1}{2}\left(đccm\right)\)
\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2017}\)
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2018}\)
\(\Rightarrow S=2S-S=1-\left(\frac{1}{2}\right)^{2018}\)
\(\Rightarrow S< 1\)( đpcm )
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2018}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(S=1-\frac{1}{2^{2018}}< 1\) ( đpcm )
Chúc bạn học tốt ~
\(C=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2017+1\)
\(=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2018-\left(2018^{2019}+2018^{2018}+...+2018\right)-1\)
\(=\left(2018^{2020}+2018^{2019}+...+2018^3+2018^2\right)-\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)+1\)\(=2018^{2020}-2018+1\)
\(=2018^{2020}-2017\)