a, S = 2 + 22 + 23 + 24 + ... + 299 + 2100. 2S = 22 + 23 + 24 + 25 + ... + 2100 + 2101 => 2S - S = S = (22 + 23 + 24 + 25 + ... + 2100 + 2101) - (2 + 22 + 23 + 24 + ... + 299 + 2100) = 2101 - 2. Vậy S = 2101 - 2. b, S = 2 + 22 + 23 + 24 + ... + 299 + 2100 = (2 + 22) + (23 + 24) + ... + (299 + 2100) = 2.(1 + 2) + 23.(1 + 2) + ... + 299.(1 + 2) = (1 + 2).(2 + 23 + ... + 299) = 3.(2 + 23 + ... + 299) => S ⋮ 3. Vậy S ⋮ 3 (đpcm)
chứng minh ý a 8mu10 cong 2 mũ 20 chia hết cho41
chứng minh ý b 31 mu36 nhận 36 -313 mu 5 nhận 299 chia hết ch
chung minhý c 3 mu n công 3 cộng 2 mu n công 3 công 2 mu n công 2 công 2 mu n công 2 chia hết 6
chung minh y d d=2 cong 2 mu 2 cong 2mu 3 cong 2 mu 4 cong 2 mu 5 cong 2 mu 6 cong ....cong 2 mu 58 cong 2 mu 59 cong 2 mu 60 chia het cho 31
chung minh y e, e= 1cong 3 cong 3 mu 2 cong 3 mu 3 cong..cong 3 mu 98 cong 3 mu 99 chung minh e chia het cho5 chia het cho11
bai 1; cho tong M =126 +213+x. Tim x de M chia het cho 3
bai 2; chung to rang tong ; A= 2 + 2 mu 3 + 2 mu 4 + 2 mu 5 + 2 mu 6 +2 mu 7 +2 mu 9 + 2 mu 10 + 2 mu 12 chia het cho 5
chung to H chia het cho 155 , biet H = 2 + 2 mu 2 + 2 mu 3 + 2 mu 4 + ......+ 2 mu 99 + 2 mu 100
chung minh rang :
a. 7^6+7^5-7^4 chia het cho 55.
b. 3 mu n cong 2 + 3 mu n cong 1tru3 mu n,chia het cho 11
chung minh rang :
a. 7^6+7^5-7^4 chia het cho 55.
b. 3 mu n cong 2 + 3 mu n cong 1tru3 mu n,chia het cho 11
cho A = 3 mu 2...3 mu 2019 . chung minh rang chia het cho 120
chung minh rang 8 mu 10 cong 2 mu 20 chia het cho 41
chung minh rang 313 mu 6 nhan 36 tru 313 mu 5 nhan 299 chia het cho 7
A= 4 + 4 mu 2 + ... + 4 mu 100
chung to rang A chia het cho 5
so sanh A va B , biet B = 4 mu 99
cho S= 1-3 + 3 mu 2 - 3 mu 3 +...+3 mu 98 - 3 mu 99
a) cmr : s la boi cua -20
b) tinh s , tu đo suy ra 3 mu 100 chia 4 du 1
