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chứng tỏ rằng : A=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+....+\frac{36}{25.27.29}< 3\)
Ta có:
\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)
\(\Rightarrow A=3-\frac{1}{87}\)
Vì \(3-\frac{1}{87}< 3.\)
\(\Rightarrow A< 3\left(đpcm\right).\)
Chúc bạn học tốt!
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{20}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}}{2^{20}}-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}-1}{2^{20}}\)
Vậy chọn câu a)
19 nha !!!!!!!!!! tick mình đi làm ơn để mình đủ 20 không mình bị phạt đấy
\(S=\frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{20}+\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{21}{20}\)
\(S=\left(\frac{3}{2}-\frac{1}{2}\right)+ \left(\frac{4}{3}-\frac{1}{3}\right)+\left(\frac{5}{4}-\frac{1}{4}\right)+...+\left(\frac{21}{20}-\frac{1}{20}\right)\)
\(S=1+1+1...+1\)
\(S=1.20=20\)
\(\frac{20}{1.3.5}+\frac{20}{3.5.7}+\frac{20}{5.7.9}+...+\frac{20}{25.27.29}\)
\(=5.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=5.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=5.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(=5.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(=5.\frac{260}{783}\)
\(=\frac{1300}{783}\)
Ta có:\(\frac{1}{\left(n-2\right)n}-\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)-\left(n-2\right)n}{\left(n-2\right)n\cdot n\left(n+2\right)}\)
\(=\frac{n\left(n+2-n+2\right)}{n\cdot\left(n-2\right)n\left(n+2\right)}=\frac{4}{\left(n-2\right)n\left(n+2\right)}\)
Áp dụng\(\frac{20}{1.3.5}+\frac{20}{3.5.7}+...+\frac{20}{25.27.29}\)
\(=5\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=5\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=5\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(=5\cdot\frac{261-1}{783}=5\cdot\frac{260}{783}=\frac{1300}{783}\)