Tính
B=\(\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+.......+\frac{3}{1+2+3+........+2015}\)
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Những câu hỏi liên quan
27 tháng 7 2017
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
15 tháng 4 2018
bạn Kiên ơi nếu biết thì giải hẳn ra nhé bạn mình cũng đang bí bài đó
19 tháng 8 2016
Ta có :
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{\left(x-1\right)x}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{\left(x-1\right)x}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x-1\right)x}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{x}\right)\)
\(=1-\frac{2}{x}\)
\(\Rightarrow1-\frac{2}{x}=\frac{2013}{2015}\)
\(\Rightarrow\frac{2}{x}=1-\frac{2013}{2015}\)
\(\Rightarrow\frac{2}{x}=\frac{2}{2015}\)
\(\Rightarrow x=2015\)
Vậy ...
11 tháng 7 2018
a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
\(B=\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+.........+\frac{3}{1+2+3+.......+2015}\)
\(=3.\left(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+......+\frac{1}{1+2+3+......+2015}\right)\)
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); \(1+2+3+4=\frac{4.5}{2}\); ............;
\(1+2+3+4+.......+2015=\frac{2015.2016}{2}\)
\(\Rightarrow B=3.\left(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+.....+\frac{1}{\frac{2015.2016}{2}}\right)\)
\(=3.\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{2015.2016}\right)\)
\(=6.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{2015.2016}\right)\)
\(=6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=6.\left(\frac{1}{2}-\frac{1}{2016}\right)=6.\frac{1007}{2016}=\frac{1007}{336}\)