Chứng minh :
A = 2 + 22 + 24 + 25 + ........+260 chia hết cho 3
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[x+1]+[x+2]+...+[x+100]=5750
[1+2+...+100].x=5750
Số các số hạng:
(100-1):1+1=100 số
=>có 100 thừa số x
Tổng trên là:
(1+100)x100:2=5050
=>x.100=5750-5050
=>x.100=700
=>x=700:100
=>x=7
Vậy x = 7
Ta có: (x + 1) + (x + 2) +...+ (x + 100) = 5750
=> (x + x + x +...... +x) + (1 + 2 + 3 + 4 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
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goi ucln (4n+3,2n+3) la d(d thuoc N*)
<=>4n+3 chia het cho d,2n+3 chia het cho d
<=>2.(2n+3)-4n+3
<=>3 chia het cho d <=>d thuoc tap hop {1;3}
do 4n va 2n chan =>2n+3 va 4n+3 ko chia het cho3
=>d=1
<=>n thuoc tap hop 1,2
vì 2+22=2+4=6 chia hết cho 3 nên 2+22+23+24+25+...+260chia hết cho 3
nên tổng A chia hết cho 3
vậy A chia hết cho 3
Giải :
A=(2 + 2^2) + ( 2^4 + 2^5 ) +........................ +(2^59 + 2^60)
A = 2* ( 1 + 2 ) + 2^4* (1+2)+.........................+2^59 *(1+2)
Ta có 1+2=3
Mà 3 chia hết cho 3 nên A chia hết cho 3