Cho B=\(2\times2^2+3\times2^3+4\times2^4+5\times2^5+...+10\times2^{10}\)
So sánh B và \(2^{14}\)
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A=\(2.2^2+3.2^3+4.2^4+...+100.2^{100}\)
\(\Rightarrow2A=2.2^3+3.2^4+4.2^5+...+100.2^{101}\)
\(\Rightarrow A-2A=2.2^2+\left(3.2^3-2.2^3\right)+\left(4.2^4-3.2^4\right)+...+\left(100.2^{100}-99.2^{100}\right)-100.2^{101}\)
\(\Rightarrow-A=2^3+\left(2^3+2^4+...+2^{100}\right)-100.2^{101}\)
Đặt \(B=\left(2^3+2^4+...+2^{100}\right)\)
\(\Rightarrow2B=\left(2^4+2^5+...+2^{101}\right)\)
\(\Rightarrow2B-B=\left(2^4+2^5+...+2^{101}\right)-\left(2^3+2^4+...+2^{100}\right)\)
\(\Rightarrow B=2^{101}-2^3\)
\(\Rightarrow-A=2^3+2^{101}-2^3-100.2^{101}\)
\(\Rightarrow-A=2^{101}-100.2^{101}\)
\(\Rightarrow A=100.2^{101}-2^{101}=99.2^{101}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
a) \(3\times\dfrac{4}{11}=\dfrac{3\times4}{11}=\dfrac{12}{11}\)
b) \(1\times\dfrac{5}{4}=\dfrac{1\times5}{4}=\dfrac{5}{4}\)
c) \(0\times\dfrac{2}{5}=\dfrac{0\times2}{5}=\dfrac{0}{5}=0\)
a: \(=\dfrac{3\cdot4}{11}=\dfrac{12}{11}\)
b: \(=\dfrac{1\cdot5}{4}=\dfrac{5}{4}\)
c: \(=\dfrac{0\cdot2}{5}=0\)
a: Số số hạng là \(\dfrac{2018-2}{2}+1=1009\left(số\right)\)
Tổng là: \(\dfrac{2018+2}{2}\cdot1009=1009\cdot1010=1019090\)
b: \(10S=10^2+10^3+...+10^{101}\)
\(\Rightarrow9S=10^{101}-10\)
hay \(S=\dfrac{10^{101}-10}{9}\)
c: \(5S=1+\dfrac{1}{5}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow4S=1-\dfrac{1}{5^{100}}\)
hay \(S=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)