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Bài 1: Tính giá trị các biểu thức:
1) \(A=\frac{2}{3}.\frac{2014}{2013}-\frac{2}{3}.\frac{1}{2013}+\frac{1}{3}\)
\(=\frac{2}{3}.\left(\frac{2014}{2013}-\frac{1}{2013}\right)+\frac{1}{3}\)
\(=\frac{2}{3}.1+\frac{1}{3}\)
= 1
A =\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+...+\(\frac{1}{1+2+3+4...+2014}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\)
\(\Rightarrow2A=2\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\right)\)
\(\Rightarrow2A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4058210}\)
\(\Rightarrow2A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{2013}{4030}\)
\(\Rightarrow A=\frac{2013}{8060}\)
ngài Kiệt ღ ๖ۣۜLý๖ۣۜ đúng là không ái sánh bằng sự gian xảo này
Bài 1:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
Bài 2:
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
Vậy A < 2
Bài 3:
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
Bài 4:
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)
A=1-1/2+1/2-1/3+.............................1/49-1/50
A=1-1/50
A=49/50
\(\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
\(=2013+\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)-\left(1+1+...+1\right)\)
\(=2013+\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}-2012\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+1\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}\)
\(=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(\Rightarrow A=\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}=2014\)
\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(P=1+\frac{1}{2}.3+\frac{1}{3}.6+\frac{1}{4}.10+....+\frac{1}{2016}.2033136\)
\(P=1+\frac{3}{2}+4+\frac{5}{2}+....+\frac{2017}{2}\)
\(P=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2017}{2}\)
\(P=\frac{2+3+4+5+....+2017}{2}=\frac{2035152}{2}=1017576\)
a. \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2014}}\right):2=\frac{1}{2}-\frac{1}{3^{2014}.2}=\frac{3^{2014}-1}{3^{2014}.2}\)
b.\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}\)
\(\Rightarrow2B=1+\frac{1}{2^2}+....+\frac{1}{2^{2013}}\)
\(\Rightarrow2B-B=1-\frac{1}{2^{2014}}\)
\(\Rightarrow B=1-\frac{1}{2^{2014}}\)
mình cũng bít làm bài đó
\(A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\)
\(2A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
\(2A-A=\left(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\right)\)
\(A=1+\frac{1}{2}-\frac{1}{2^{2014}}\)
\(A=\frac{2^{2014}}{2^{2014}}+\frac{2^{2013}}{2^{2014}}-\frac{1}{2^{2014}}\)
\(A=\frac{2^{2014}+2^{2013}-1}{2^{2014}}\)
Tham khảo nhé~