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\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+...+\dfrac{1}{2013.2015.2017}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...+\dfrac{4}{2013.2015.2017}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}-\dfrac{1}{2015.2017}\right)\)\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{2015.2017}\right)=\dfrac{1}{12}-\dfrac{1}{4.2015.2017}\)
ta có
\(A=6\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+..+\frac{4}{25.27.29}\right)=6\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+..+\frac{29-25}{25.27.29}\right)\)
\(=6\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+..+\frac{1}{25.27}-\frac{1}{27.29}\right)=6\left(\frac{1}{3}-\frac{1}{27.29}\right)\)
\(=2-\frac{2}{9.29}=\frac{520}{261}\)
Đặt tổng là A
\(\frac{A}{6}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{6}{25.27.29}\)
\(\frac{A}{6}=\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{29-25}{25.27.29}\)
\(\frac{A}{6}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\frac{A}{6}=\frac{1}{1.3}-\frac{1}{27.29}\Rightarrow A=\left(\frac{1}{3}-\frac{1}{27.29}\right):6\)
\(A=\frac{24}{1.3.5}+\frac{24}{3.5.7}+\frac{24}{5.7.9}+...+\frac{24}{25.27.29}\)
\(A=6\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(A=6\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{29-25}{25.27.29}\right)\)
\(A=6\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(A=6\left(\frac{1}{1.3}-\frac{1}{27.29}\right)=\frac{520}{261}\)
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!
Cho x và y thoả mãn (x-45)2=-|2y+5| tính giá trị của biểu thức: M=x2+y2+29/10.y-15
1/1.3.5 + 1/3.5.7 + 1/5.7.9 +.....+ 1/99.101.103
= 1/4. [4/1.3.5 + 4/3.5.7 + 4/ 5.7.9 +....+ 4/99.101.103]
=1/4. [1/1.3 - 1/3.5 + 1/3.5 - 1/5.7 +....+ 1/99.101 - 1/101.103]
= 1/4. [1/1.3 - 1/101.103]
=1/4. 10406/31209
= 5230/62418
\(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+....+\frac{1}{99\cdot101\cdot103}\)
\(2A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5-7}+....+\frac{1}{99\cdot101}-\frac{1}{101\cdot103}\)
\(2A=\frac{1}{1\cdot3}-\frac{1}{101\cdot103}\)
Tính nốt