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Ta có :
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{30}{40}=\frac{3}{4}\)
\(\Rightarrow\)\(A>\frac{3}{4}\) ( điều phải chứng minh )
Vậy \(A>\frac{3}{4}\)
Chúc bạn học tốt ~
a) C=\(\left(1+3+3^2\right)+....+\left(3^9+3^{10}+3^{11}\right)\)
=13+.....+3^11 chia het cho 13
nen C=1+3+...+3^11 chia het cho 13
\(C=1+3+3^2+3^3+......+3^{11}\)
\(C=\left(1+3+3^2\right)+.......+\left(3^9+3^{10}+3^{11}\right)\)
\(C=13.\left(1+3+3^2\right)+........+13.\left(1+3+3^2\right)\)
Mà 13 \(⋮\)13 => C \(⋮\)13
Tương tự với câu b
b) \(C=1+3+3^2+3^3+.......+3^{11}\)
\(C=\left(1+3+3^2+3^3\right)+......+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(C=40.\left(1+3+3^2+3^3\right)+......+40.\left(1+3+3^2+3^3\right)\)
Mà 40 \(⋮\)40 => C \(⋮\)40
a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
1/A=1.21.22.23.24.25 câu 2 làm tương tự
A.2=2.22.23.24.25.26
A.2-A=(2.22.23.24.25.2 mũ 6)-(1.21.22.23.24.25)
A=26-1
3 A=1+3+32+33+...37
3.A=3+32+33+34...+38
2A=38-1
A=(38-1):2
\(a)\) Đề sai nhé
Ta có :
\(A=1+3+3^2+...+3^{11}\)
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\)
\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^9\left(1+3+3^2\right)\)
\(A=\left(1+3+9\right)+3^3\left(1+3+9\right)+...+3^9\left(1+3+9\right)\)
\(A=13+3^3.13+...+3^9.13\)
\(A=13\left(1+3^3+...+3^9\right)⋮13\)
Vậy \(A⋮13\)
Chúc bạn học tốt ~
1)
\(A=156+273+533+y\)
\(A=962+y\)
\(962⋮13\)
Để \(A⋮13\rightarrow y⋮13\)
\(A⋮̸13\rightarrow y⋮̸13\)
2)
\(A=1+3+3^2+...+3^{11}\)
* để A chia hết cho 13:
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\)
\(A=1\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^9\left(1+3+3^2\right)\)
\(A=\left(1+3^3+...+3^9\right)\left(1+3+3^2\right)\)
\(A=13\left(1+3^3+3^9\right)⋮13\rightarrowđpcm\)
* để A chia hết cho 40:
\(A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(A=1\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^8\left(1+3+3^2+3^3\right)\)\(A=\left(1+3^4+...+3^8\right)\left(1+3+3^2+3^3\right)\)
\(A=40\left(1+3^4+...+3^8\right)⋮40\rightarrowđpcm\)
3)
\(25^{24}-25^{23}\)
\(=25^{23}.25-25^{23}.1\)
\(=25^{23}.\left(25-1\right)\)
\(=25^{23}.24\)
\(=25^{23}.4.6⋮6\rightarrowđpcm\)
4) Gọi 5 số tự nhiên liên tiếp đó là a;a+1;a+2;a+3;a+4
Tích của 5 số tự nhiên liên tiếp là :
\(a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\)
Ta có: \(a+1;a+3\) hoặc \(a+2;a+4\)là 2 số chẵn liên tiếp nên sẽ chia hết cho 8
5 số tự nhiên liên tiếp luôn có 1 số chia hết cho 5
a;a+1;a+2 luôn sẽ có 1 số chia hết cho 3
5 số tự nhiên liên tiếp đó chia hết cho 3;5;8
\(\Rightarrow⋮120\rightarrowđpcm\)
Bài 1:
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow\dfrac{1}{5}A=\dfrac{1}{5^2}+\dfrac{1}{5^3}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{100}}\)
Lây vế trừ vế, ta được:
\(A-\dfrac{1}{5}A=\dfrac{4}{5}A\)
\(\dfrac{4}{5}A=\dfrac{1}{5}-\dfrac{1}{5^{100}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{5}-\dfrac{1}{5^{100}}}{\dfrac{4}{5}}=\dfrac{\dfrac{1}{5}.\left(1-\dfrac{1}{5^{99}}\right)}{\dfrac{1}{5}.4}=\dfrac{1-\dfrac{1}{5^{99}}}{4}\)
Vậy \(A=\dfrac{1-\dfrac{1}{5^{99}}}{4}\).
Chúc bạn học tốt!
Bài 2:
Có:
\(B=3+3^3+3^5+...+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+...+3^{1986}\left(3+3^3+3^5\right)\)
\(\Leftrightarrow B=273+...+3^{1986}.273\)
\(\Leftrightarrow B=273\left(1+...+1986\right)\)
Vì \(273⋮13\)
Nên \(B=273\left(1+...+1986\right)⋮13\)
Vậy \(B⋮13\)
Lại có:
\(B=3+3^3+3^5+...+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+...+3^{1984}\left(3+3^3+3^5+3^7\right)\)
\(\Leftrightarrow B=2460+...+3^{1984}.2460\)
\(\Leftrightarrow B=2460\left(1+...+3^{1984}\right)\)
Vì \(2460⋮41\)
Nên \(B=2460\left(1+...+3^{1984}\right)⋮41\)
Vậy \(B⋮41\).
Chúc bạn học tốt!
Lời giải:
Ta có:
\(A=1+3+3^2+3^3+...+3^{11}\)
\(=(1+3+3^2)+(3^3+3^4+3^5)+....+(3^9+3^{10}+3^{11})\)
\(=(1+3+3^2)+3^3(1+3+3^2)+...+3^9(1+3+3^2)\)
\(=(1+3+3^2)(1+3^3+...+3^9)=13(1+3^3+...+3^9)\vdots 13\) (đpcm)
Và:
\(A=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^{10}+3^{11})\)
\(=(1+3+3^2+3^3)+3^4(1+3+3^2+3^3)+3^8(1+3+3^2+3^3)\)
\(=(1+3+3^2+3^3)(1+3^4+3^8)=40(1+3^4+3^8)\vdots40\)