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![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)
b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)
c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)
\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)
\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)
Tương tự câu d,e,f bạn tự làm nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a,19/7=5/7 +2
2>7/9 => 19/7>7/9
b, 72/73=1- 1/73
98/99=1- 1/99
1/73>1/99
c,19/18=1+ 1/18
2005/2004=1+ 1/2004
1/18>1/2004
d, 72/73=(58+14)/73=58/73 + 14/73
58/73>58/99
=> 72/73>58/99
![](https://rs.olm.vn/images/avt/0.png?1311)
F = 7 + 72 + 73 + 74 + ..... + 7100
F= 7+(1+7)+73+(1+7)+...+799+(1+7)
F = 7x8+73x8+...+799x8
F= 8x(7+73+...+799)
mà 8 chia hết 8 => 8(7+73+...+799) chia hết 8
Vậy F chia hết cho 8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=7+7^2+7^3+...7^{20}\)
Ta có: \(7S=7.\left(7+7^2+7^3+...+7^{20}\right)\)
\(7S=7^2+7^3+7^4+...+7^{21}\)
\(7S-S=\left(7^2+7^3+7^4+...+7^{21}\right)-\left(7+7^2+7^3+...+7^{20}\right)\)
\(6S=\left(7^{21}-7\right)\)
\(S=\left(7^{21}-7\right):6\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=1+7+7^2+7^3+...+7^{150}\)
\(7B=7.\left(1+7+7^2+7^3+...+7^{150}\right)\)
\(7B=7+7^2+7^3+7^4+...+7^{151}\)
\(7B-B=\left(7+7^2+7^3+7^4+...+7^{151}\right)-\left(1+7+7^2+7^3+...+7^{150}\right)\)
\(6B=\left(7^{151}-1\right)\)
\(B=\left(7^{151}-1\right):6\)
B = 1 + 7 + 72 + ...+ 7150
7.B = 7 + 72+.....+ 7150 + 7151
7B - B = 7151 - 1
6B = 7151 - 1
B = \(\dfrac{7^{151}-1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta xét biểu thức \(A_1=7+7^2+7^3\) \(=7\left(1+7+7^2\right)\) \(=57.7⋮57\)
\(A_2=7^4+7^5+7^6\) \(=7^4\left(1+7+7^2\right)\) \(=57.7^4⋮57\)
...
\(A_{40}=7^{118}+7^{119}+7^{120}\) \(=7^{118}\left(1+7+7^2\right)⋮57\)
Vậy \(A=\sum\limits^{40}_{i=1}A_i\) đương nhiên chia hết cho 57 (đpcm)
D=1+7+72+...+7300
7D=7+72+73+...+7301
7D-D=(7+72+73+...+7301)-(1+7+72+...+7300)
6D=7301-1
=>D=\(\frac{7^{301}-1}{6}\)
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