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NN
28 tháng 11 2017
1)
a) \(A=3+3^2+3^3+3^4+3^5+3^6+....+3^{28}+3^{29}+3^{30}\)
\(\Leftrightarrow A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{28}+3^{29}+3^{30}\right)\)
\(\Leftrightarrow A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+....+3^{28}\left(1+3+3^2\right)\)
\(\Leftrightarrow A=3.13+3^4.13+....+3^{28}.13\)
\(\Leftrightarrow A=13\left(3+3^4+....+3^{28}\right)⋮13\left(dpcm\right)\)
b) \(A=3+3^2+3^3+3^4+3^5+3^6+....+3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}\)
\(\Leftrightarrow A=\left(3+3^2+3^3+3^4+3^5+3^6\right)+....+\left(3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}\right)\)
\(\Leftrightarrow A=3\left(1+3+3^2+3^3+3^4+3^5\right)+....+3^{25}\left(1+3+3^2+3^3+3^4+3^5\right)\)
\(\Leftrightarrow A=3.364+....+3^{25}.364\)
\(\Leftrightarrow A=364\left(3+3^5+3^{10}+....+3^{25}\right)\)
\(\Leftrightarrow A=52.7\left(3+3^5+3^{10}+....+3^{25}\right)⋮52\left(dpcm\right)\)
NN
28 tháng 11 2017
2) \(A=3+3^2+3^3+....+3^{30}\)
\(\Leftrightarrow3A=3\left(3+3^2+3^3+....+3^{30}\right)\)
\(\Leftrightarrow3A=3^2+3^3+3^4+....+3^{30}+3^{31}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+....+3^{30}+3^{31}\right)-\left(3+3^2+3^3+....+3^{30}\right)\)
\(\Leftrightarrow2A=3^{31}-3\)
\(\Leftrightarrow A=\dfrac{3^{31}-3}{2}\)
Vậy A không phải là số chính phương
23 tháng 6 2018
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
DL
8 tháng 7 2018
1) \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(5P=\frac{1}{5^1}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
\(5P-P=\frac{1}{5^1}+\left(\frac{2}{5^2}-\frac{1}{5^2}\right)+\left(\frac{3}{5^3}-\frac{2}{5^3}\right)+...+\left(\frac{11}{5^{11}}-\frac{10}{5^{11}}\right)-\frac{11}{5^{12}}\)
\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(5A-A=1+\frac{1}{5}-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(4A=1-\frac{1}{5^{11}}\Rightarrow A=\frac{1-\frac{1}{5^{11}}}{4}\)
\(4P=\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}=\frac{1-\frac{1}{5^{11}}}{16}-\frac{11}{5^{12}\cdot4}< \frac{1}{16}\)
NT
26 tháng 12 2017
1. \(A=2^{2016}-1\)
\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)
\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)
16 chia 5 dư 1 nên 16^504 chia 5 dư 1
=> 16^504-1 chia hết cho 5
hay A chia hết cho 5
\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)
lý luận TT trg hợp A chia hết cho 5
(3;5;7)=1 = > A chia hết cho 105
2;3;4 TT ạ !!
12 tháng 12 2016
b, A = 3+3^2 +3^3 +3^4 +....+3^120 =﴾3+3^2+3^3﴿+......+﴾3^118+3^119+3^120﴿ =3﴾1+3+3^2﴿+....+3^118﴾1+3+3^2﴿ = 3.13+...+3^118. 13 = 13﴾ 3+...+3^118﴿ chia hết cho 13 c, A = 3+3^2 +3^3 + 3^4 +....+3^120 = ﴾3+3^2+3^3+3^4﴿+.....+﴾3^117+3^118+3^119+3^120﴿ = 3﴾1+3+3^2+3^3﴿ +...+3^117﴾ 1+3+3^2 +3^3﴿ = 3.40+ ...+3^117 .40 = 40 .﴾ 3+....+3^117﴿ chia hết cho 40
12 tháng 12 2016
b, A = 3+3^2 +3^3 +3^4 +....+3^120
=(3+3^2+3^3)+......+(3^118+3^119+3^120)
=3(1+3+3^2)+....+3^118(1+3+3^2)
= 3.13+...+3^118. 13
= 13( 3+...+3^118) chia hết cho 13
c, A = 3+3^2 +3^3 + 3^4 +....+3^120
= (3+3^2+3^3+3^4)+.....+(3^117+3^118+3^119+3^120)
= 3(1+3+3^2+3^3) +...+3^117( 1+3+3^2 +3^3)
= 3.40+ ...+3^117 .40
= 40 .( 3+....+3^117) chia hết cho 40
DN
19 tháng 5 2021
b
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+..+\frac{1}{70}\)
Ta thấy:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( có 10 phân số \(\frac{1}{20}\)) = \(\frac{1}{20}\).10 = \(\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 phân số \(\frac{1}{30}\)) = \(\frac{1}{30}\).10 = \(\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( có 10 phân số \(\frac{1}{40}\)) = \(\frac{1}{40}\).10 = \(\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)( có 10 phân số \(\frac{1}{50}\)) =\(\frac{1}{50}.10=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( có 10 phân số \(\frac{1}{60}\)) =\(\frac{1}{60}.10=\frac{1}{6}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\)( có 10 phân số \(\frac{1}{70}\)) \(=\frac{1}{70}.10=\frac{1}{7}\)
=> A> \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{223}{140}=\frac{699}{420}>\frac{560}{420}=\frac{4}{3}\)
=> A > \(\frac{4}{3}\)
\(A=3+3^2+..+3^{24}\\ =\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{23}+3^{24}\right)\\ =12+3^2\left(3+3^2\right)+....+3^{22}\left(3+3^2\right)\\ =12+3^2.12+...+3^{22}.12\\ =12\left(1+3^2+...+3^{22}\right)⋮12\)
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