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Theo đề bài ta có:
A = \(1+2+2^2+2^3+...+2^{11}\)
\(\Rightarrow A=2^0+2^1+2^2+2^3+...+2^{11}\)
\(\Leftrightarrow A=2^0.\left(1+2+2^2+2^3+2^4+2^5\right)+2^6.\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(\Rightarrow A=2^0.63+2^6.63\)
\(\Rightarrow A=63.\left(2^0+2^6\right)\)
\(\Rightarrow A=63.65\)
Vậy A chia hết cho 13 ( vì 65 chia hết cho 13)
a) \(4^{13}+4^{14}+4^{15}+4^{16}=4^{13}\left(1+4\right)+4^{14}\left(1+4\right)=4^{13}.5+4^{14}.5=5\left(4^{13}+4^{14}\right)⋮5\Rightarrow dpcm\)
c) \(2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}\)
\(=2^{10}\left(1+2+2^2\right)+2^{13}\left(1+2+2^2\right)\)
\(=2^{10}.7+2^{13}.7=7\left(2^{10}+2^{13}\right)⋮7\Rightarrow dpcm\)
Câu c bạn xem lại đê
3/10>3/15
3/11>3/15
3/12>3/15
3/13>3/15
3/14>3/15
=>S>3/15*5=15/15=1
3/11<3/10
3/12<3/10
3/13<3/10
3/14<3/10
=>3/11+3/12+3/13+3/14+3/10<3/10*5=15/10=3/2<2
=>1<S<2
a) \(A=3+3^2+..+3^{60}\)
\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{59}\cdot\left(1+3\right)\)
\(A=4\cdot\left(3+3^3+...+3^{59}\right)\)
Vậy A chia hết cho 4
b) \(A=3+3^2+3^3+...+3^{60}\)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(A=3\cdot\left(1+3+3^2\right)+...+3^{58}\cdot\left(1+3+3^2\right)\)
\(A=13\cdot\left(3+..+3^{58}\right)\)
Vậy A chia hết cho 13
Sửa lại đề nha:
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{15}{14}>1\left(1\right)\)
Lại có:
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow1< S< 2\left(đpcm\right)\)
Chúc em học tốt!
S=3/10+3/11+3/12+3/13+3/14 > 3/14+3/14+3/14+3/14+3/14=15/14>1
=>3/10+3/11+3/12+3/13+3/14>1=>S>1
S=3/10+3/11+3/12+3/13+3/14 < 3/11+3/11+3/11+3/11+3/11=15/11<2
=>3/10+3/11+3/12+3/13+3/14<2=>S<2
\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)
\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{9}+\frac{3}{9}+\frac{3}{9}+\frac{3}{9}+\frac{3}{9}=\frac{15}{9}< \frac{18}{9}=2\)
Suy ra đpcm.