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a) Rút gọn : \(M=5+5^2+5^3+...+5^{100}\)
b) Chứng tỏ : \(N=5^1+5^2+5^3+5^4+...+5^{2010}⋮6\) và \(31\)
a, \(M=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5M=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5M-M=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+....+5^{100}\right)\)
\(\Rightarrow4M=5^{101}-5\)
\(\Rightarrow M=\frac{5^{101}-5}{4}\)
Vậy : \(M=\frac{5^{101}-5}{4}\)
Ta có :
\(N=5+5^2+5^3+....+5^{2010}\)
\(\Rightarrow N=5\left(1+5+5^2\right)+.....+5^{2008}\left(1+5+5^2\right)\)
\(\Rightarrow N=5.31+....+2^{2008}.31\)
=> N chia hết cho 31
\(N=5^1+5^2+5^3+5^4+...+5^{2010}\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{2018}\left(1+5+5^2\right)\)
\(=31\left(5+5^4+...+5^{2018}\right)⋮31\)
=>đpcm
Bài 1
a) A = 2^0 + 2^1 + 2^2 +...+ 2^50
2A=2^1+2^2+2^3+...+2^51
2A-A=(2^1+2^2+2^3+...+2^51)-(2^0 + 2^1 + 2^2 +...+ 2^50)
A=(2^1-2^1)+(2^2-2^2)+...+(2^50-2^50)+(2^51-2^1)
A=0+0+...+0+(2^51-2^1)
A=2^51-2^1
b)B = 5 + 5^2 + 5^3 +...+ 5^99 + 5^100
5B=5^2+5^3+5^4+...+5^100+5^101
5B-B=(5^2+5^3+5^4+...+5^100+5^101)-( 5 + 5^2 + 5^3 +...+ 5^99 + 5^100)
4B=(5^2-5^2)+(5^3-5^3)+...+(5^100-5^100)+(5^101-5)
4B=0+0+...+0+(5^101-5)
4B=5^101-5
B=(5^101-5)/4
c)C = 3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010
3C=3^2-3^3+3^4-3^5+...+3^2010-3^2011
3C-C=(3^2-3^3+3^4-3^5+...+3^2010-3^2011)-(3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010)
...............................................!!!!!!!!!!!!!!!!!!!!!!!!
Bài 2
8(mình k0 chắc)
A=2^1+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)
=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)
=2.3+2^3.3+...+2^2010.3
=(2+2^3+2^2010).3
=> A chia het cho 3
Bài 1:
$A=2^1+2^2+2^3+2^4$
$2A=2^2+2^3+2^4+2^5$
$\Rightarrow 2A-A=2^5-2^1$
$\Rightarrow A=2^5-1=32-1=31$
----------------------------
$B=3^1+3^2+3^3+3^4$
$3B=3^2+3^3+3^4+3^5$
$\Rightarrow 3B-B = 3^5-3$
$\Rightarrow 2B = 3^5-3\Rightarrow B = \frac{3^5-3}{2}$
--------------------------
$C=5^1+5^2+5^3+5^4$
$5C=5^2+5^3+5^4+5^5$
$\Rightarrow 5C-C=5^5-5$
$\Rightarrow C=\frac{5^5-5}{4}$
( 21 + 22 ) + ( 23 + 24 ) + ... + ( 22009 + 22010 )
= 2. ( 1 + 2 ) + 23 . ( 1 + 2 ) + ... + 22009 . ( 1 + 2 )
= 3 . ( 2 + 23 + ... + 22009 ) chia hết cho 3. => ĐPCM
\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)
\(2A=2+2^2+2^3+...+2^{51}\)
\(2A-A=A=2^{51}-2^0\)
\(B=5+5^2+5^3+...+5^{99}+5^{100}\)
\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=4B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)
\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)
\(3C+C=4C=3^{2011}+3\)
\(C=\frac{3^{2011}+3}{4}\)
\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)
\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)
\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)
\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)
\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)
a) \(M=5+5^2+5^3+...+5^{100}\)
=> \(5M=\left(5+5^2+5^3+...+5^{100}\right).5\)
= \(5^2+5^3+5^4+...+5^{101}\)
=> \(5M-M=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
=> \(4M=5^{101}-5\)
=> \(M=\frac{5^{101}-5}{4}\)