NQ

Cho M= 53+832+1133+3023100CMR M

H9
2 tháng 7 2024 lúc 8:21

\(M=\dfrac{5}{3}+\dfrac{8}{3^2}+...+\dfrac{302}{3^{100}}\\ 3M=5+\dfrac{8}{3}+\dfrac{11}{3^2}+...+\dfrac{302}{3^{99}}\\ 3M-M=\left(5+\dfrac{8}{3}+\dfrac{11}{3^2}+...+\dfrac{302}{3^{99}}\right)-\left(\dfrac{5}{3}+\dfrac{8}{3^2}+...+\dfrac{302}{3^{100}}\right)\\ 2M=5+\left(\dfrac{8}{3}-\dfrac{5}{3}\right)+\left(\dfrac{11}{3^2}-\dfrac{8}{3^2}\right)+....+\left(\dfrac{302}{3^{99}}-\dfrac{299}{3^{99}}\right)+\dfrac{302}{3^{100}}\\ 2M=5+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{302}{3^{100}}\\ 2M=6+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{302}{3^{100}}\\ 6M=18+1+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}+\dfrac{302}{3^{99}}\\ 6M-2M=\left(19+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}+\dfrac{302}{3^{99}}\right)-\left(6+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{302}{3^{100}}\right)\\ 4M=13-\dfrac{1}{3^{98}}+\dfrac{302}{3^{99}}-\dfrac{302}{3^{100}}\\ 4M=13+\dfrac{595}{3^{100}}\\ M=\dfrac{1}{4}\left(13+\dfrac{595}{3^{100}}\right)=\dfrac{13}{4}+\dfrac{1}{4}\cdot\dfrac{595}{3^{100}}\)  

Vì: 

\(595< 3^{100}=>\dfrac{595}{3^{100}}< 1=>\dfrac{1}{4}\cdot\dfrac{595}{3^{100}}< \dfrac{1}{4}\\ =>M=\dfrac{13}{4}+\dfrac{1}{4}\cdot\dfrac{595}{3^{100}}< \dfrac{13}{4}+\dfrac{1}{4}=\dfrac{7}{2}=3\dfrac{1}{2}\)

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