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a) S = 5 + 52 + 53 + ... + 5100
=> S = ( 5 + 52 ) + ( 53 + 54 ) + ... + ( 599 + 5100 )
=> S = 5( 1 + 5 ) + 53( 1 + 5 ) + ... + 599( 1 + 5 )
=> S = 5 . 6 + 53 . 6 + ... + 599 . 6
=> S = ( 5 + 53 + ... + 599 ) . 6 chia hết cho 6
=> S chia hết cho 6
b) S1 = 2 + 22 + 23 + ... + 2100
=> S1 = ( 2 + 22 + 23 + 24 + 25 ) + ... + ( 296 + 297 + 298 + 299 + 2100 )
=> S1 = 2( 1 + 2 + 22 + 23 + 24 ) + ... +296( 1 + 2 + 22 + 23 + 24 )
=> S1 = 2 . 31 + ... + 296 . 31
=> S1 = ( 2 + ... + 296 ) . 31 chia hết cho 31
=> S1 chia hết cho 31
c) S2 = 165 + 215
=> S2 = ( 24 )5 + 215
=> S2 = 220 + 215
=> S2 = 220( 1 + 25 )
=> S2 = 220 . 33 chia hết cho 33
=> S2 chia hết cho 33
\(S_2=2+2^2+2^3+2^4+.........+2^{99}+2^{100}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+.....+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(2+2^2+2^3+2^4\right)+2^5\left(2+2^2+2^3+2^4\right)+......+2^{97}\left(2+2^2+2^3+2^4\right)\)
\(=2.31+2^5.31+......+2^{97}.31\)
\(=31\left(2+2^5+....+2^{97}\right)⋮31\left(đpcm\right)\)
c. S3 = 165 + 215 chia hết cho 33
ta thấy: 16^5=2^20
=> A=16^5 + 2^15 = 2^20 + 2^15
= 2^15.2^5 + 2^15
= 2^15(2^5+1)
=2^15.33
số này luôn chia hết cho 33
b. S2 = 2 + 22 + 23 + 24 +........... + 2100 chia hết cho 31
= 2(1 + 2 + 22 + 23 + 24 ) + 26( 1 + 2 + 22 + 23 + 24 ) + ....+ (1 + 2 + 22 + 23 + 24 )296
= 2 x 31 + 26 x 31 + ..... + 296 x 31 = 31 x ( 2 + 26 + ..... + 296 )
=> 2 + 22 + 23 + 24 +........... + 2100 chia hết cho 31
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
b)\(2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+....+2^{96}.31\)
\(=31.\left(2+....+2^{96}\right)⋮31\)
Vậy...
a) \(5+5^2+5^3+...+5^{2004}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...\left(5^{2003}+5^{2004}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2003}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{2003}.6\)
\(=6.\left(5+5^3+...+5^{2003}\right)⋮6\)
Vậy....
\(5+5^2+5^3+...+5^{2004}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6+\right)+...+\left(5^{2002}+5^{2003}+5^{2004}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{2002}\left(1+5+5^2\right)\)
\(=5.31+5^4.31+...+5^{2002}.31\)
\(=31.\left(5+5^4+...+5^{2002}\right)⋮31\)
Vậy...
Trường hợp 3 làm tương tự để chứng minh
a)Ta có:S1=5+52+53+…+599+5100
=>5.S1=52+53+54+…+5100+5101
=>5.S1-S1=52+53+54+…+5100+5101-5-52-53-…-599-5100
=>4.S1=5101-5
=>\(S_1=\frac{5^{101}-5}{4}\)
b)S2=2+22+23+…+299+2100
=>2.S2=22+23+24+…+2100+2101
=>2.S2-S2=22+23+24+…+2100+2101-2-22-23-…-299-2100
=>S2=2101-2
2S1=52+53+54+....+5100+5101
2S1-s1=5101-5
S1=5101-5
b) S2=2101-2
Ta có: 5/22 < 5/1.2
5/32 < 5/2.3
....
5/1002 < 5/99.100
⇒ S < 5/1.2 + 5/2.3 + 5/3.4+....+ 5/99.100
S< 5. (1/1.2+ 1/2.3+ ... + 1/99.100)
S< 5. (1 - 1/2 + 1/2 -1/3 + ...+1/99 - 1/100)
S< 5. (1-1/100)
S< 5.99/100
S< 4.95< 5
⇒S< 5
Vậy S<5