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Có :
2A = 1+1/2+1/2^2+.....+1/2^2015
A = 2A - A = (1+1/2+1/2^2+......+1/2^2015) - (1/2+1/2^2+.....+1/2^2016)
= 1 - 1/2^2016
Tk mk nha
Xét \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
Trừ vế theo vế ta được:
\(2A-A=A=1-\frac{1}{2^{2016}}\)
Không thể tính được nữa vì số mũ quá lớn!!!
b) \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}\)
\(B=1-\frac{1}{2015}\)
\(B=\frac{2014}{2015}\)
a) \(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
b)\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}\)
\(=\frac{2014}{2015}\)
còn lại tự giải nha gần giống như phần b thôi cũng thú vị.
ủng hộ nha
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(A=2.\frac{3}{16}\)
\(A=\frac{3}{8}\)
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(B=\frac{1}{3}-\frac{1}{21}\)
\(B=\frac{2}{7}\)
Bài 1:
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) ta có: \(A=1+2+2^2+2^3+...+2^{2018}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2019}\)
\(\Rightarrow2A-A=2^{2019}-2\)
\(\Rightarrow A=2^{2019}-2\)
Chúc bn học tốt !!!!!
a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)