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a/M=2/3.5+2/5.7+2/7.9+.....+2/97.99
M=1/3-1/5+1/5-1/7+..+1/97-1/99
M=1/3-1/99
M=32/99
b)ta có 1/2.3+1/3.4+1/4.5+..+1/2015.2016+1/2016.2017<A
=>1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016+1/2016-1/2017<a
1/2-1/2017<A
2/15/4034<A (1)
Ta có
1/1.2+1/2.3+1/3.4+1/4.5+..+1/2015.2016>A
=>1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016>A
1-1/2016
2015/2016>A (2)
Từ (1) và (2)=>A không phải là số tự nhiên(đpcm)
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}\Leftrightarrow D< 1\left(đpcm\right)\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\)
Ta thấy:
\(\dfrac{1}{2^2}>0\)
\(\dfrac{1}{3^2}>0\)
\(\dfrac{1}{4^2}>0\)
...
\(\dfrac{1}{2016^2}>0\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}>2015\cdot0=0\\ \Leftrightarrow A>0\)
Mặt khác:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=\dfrac{1}{1}-\dfrac{1}{2} \)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{2016^2}< \dfrac{1}{2015\cdot2016}=\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\\ \Leftrightarrow A< 1-\dfrac{1}{2016}< 1\left(2\right)\)Từ (1) và (2) ta có: \(0< A< 1\)
Không có số tự nhiên nào nằm giữa 0 và 1, vậy A không phải là số tự nhiên
Bài 1:
a) Đặt A = 1 + 7 + 72 + 73 + ... + 72016
7A = 7 + 72 + 73 + 74 + ... + 72017
7A - A = (7 + 72 + 73 + 74 + ... + 72017) - (1 + 7 + 72 + 73 + ... + 72016)
6A = 72017 - 1
\(A=\frac{7^{2017}-1}{6}\)
b) Đặt B = 1 + 4 + 42 + 43 + ... + 42017
4B = 4 + 42 + 43 + 44 + ... + 42018
4B - B = (4 + 42 + 43 + 44 + ... + 42018) - (1 + 4 + 42 + 43 + ... + 42017)
3B = 42018 - 1
\(B=\frac{4^{2018}-1}{3}\)
Bài 2:
a) Ta có: \(14\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}-1⋮13\left(đpcm\right)\)
b) Ta có: \(2015\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}-1⋮2014\left(đpcm\right)\)
Sorry mình thiếu 1+7+72+73+...+72016 câu dưới cũng thiếu 4 nha
12+22+32+..........+20132+20142+20152 Gọi dãy trên là A
=1x1+2x2+3x3+.........+2013x2013+2014x2014+2015x2015
=1x(2-1)+2x(3-1)+3x(4-1)+........+2013x(2014-1)+2014x(2015-1)+2015x(2016-1)
=1x2-1x1+2x3-2x1+3x4-3x1+......+2013x2014-2013x1+2014x2015-2014x1+2015x2016-2015x1
=(1x2+2x3+3x4+.........+2013x2014+2014x2015+2015x2016)-(1+2+3+........+2013+2014+2015)
Gọi vế 1 của dãy là a
3xa=1x2x3+2x3x(4-1)+3x4x(5-2)+......+2013x2014x(2015-2012)+2014x2015x(2016-2013)+2015x2016x(2017-2014)
3xa=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+........+2013x2014x2015-2012x2013x2014+2014x2015x2016-2013x2014x2015+2015x2016x2017-2014x2015x2016
a=2015x2016x2017:3
a=2731179360
A=2731179360-(1+2+3+.....+2013+2014+2015)
A=2731179360-[2015x(2015+1):2]
A=2731179360-2031120
A=2729143240
Nhớ tick cho mình nha
\(M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+.....+\frac{2}{97}-\frac{2}{99}\)
\(M=\frac{2}{3}-\frac{2}{99}=\frac{64}{99}\)
\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot..\cdot\left(\frac{1}{10^2}-1\right)\)
\(=\left(\frac{1}{2}\cdot\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}\cdot\frac{1}{3}-1\right)\cdot...\cdot\left(\frac{1}{10}\cdot\frac{1}{10}-1\right)\)
\(=\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot...\cdot\frac{-99}{100}\)
\(=\frac{\left(-1\right).\left(-3\right)}{2.2}\cdot\frac{\left(-2\right).\left(-4\right)}{3.3}\cdot...\cdot\frac{\left(-9\right).\left(-11\right)}{10.10}\)
\(=\frac{\left(-1\right).\left(-2\right)....\left(-9\right)}{2.3....10}\cdot\frac{\left(-3\right).\left(-4\right)....\left(-11\right)}{2.3.....10}\)
\(=\frac{-1}{10}\cdot\frac{-11}{2}=\frac{-11}{20}\)