Tính:
a) A= 2100 - 299 - 298 -....- 2
b) B= (1-\(\dfrac{1}{2}\)) . (1-\(\dfrac{1}{3}\)) . (1-\(\dfrac{1}{4}\) ) .........(1-\(\dfrac{1}{2018}\))
c) C= (1+\(\dfrac{1}{2}\)) + (1+\(\dfrac{1}{3}\)) + (1+\(\dfrac{1}{4}\)) +.....+ (1+\(\dfrac{1}{2018}\))
d) \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{1}{99.100}\)
e) \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + \(\dfrac{1}{7.9}\) +....+ \(\dfrac{1}{97.99}\)
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b) \(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2018}\right)\)
\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}....\frac{2018-1}{2018}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1.2.3...2017}{2.3.4...2018}=\frac{1}{2018}\)
c) Giữa các biểu thức là dấu nhân hay dấu cộng vậy bạn?
d)
\(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(D=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
e) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)
\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow E=\frac{16}{99}\)
a)
\(A=2^{100}-2^{99}-2^{98}-....-2\)
\(\Rightarrow 2A=2^{101}-2^{100}-2^{99}-....-2^2\)
Lấy 2 vế trừ đi lại :
\(2A-A=(2^{101}-2^{100}-...-2^2)-(2^{100}-2^{99}-...-2)\)
\(=2^{101}-2^{100}-2^{100}+2=2^{101}-2.2^{100}+2=2\)