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Mik làm được 1 bài thôi . Tiếc quá mình sắp phải đi học rồi.
BÀi 12:
S=1 + 2 + 22 + `23 +..........+ 22017
2S=2 + 22 + `23 + 24 +..........+22017 + 22018
Trừ đi hai vế ta được:
S=1 + 22018
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M = (2+22+23+24)+(25+26+27+28)+....+(217+218+219+220)
= 2.(1+2+22+23)+2^5.(1+2+22+23)+.....+217.(1+2+22+23)
= 2.15 + 25.15 +.... +217.15
= 15.(2+25+....+217) \(⋮\)15
M = \(\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...\)\(+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
= \(30+2^4\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\)
=\(30+2^4.30+...+2^{16}.30\)
=\(30\left(1+2^4+...+2^{16}\right)\)\(⋮15\)
Vậy....
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ta có;
3-3^2=-3
3^3-3^4=-3
....... 3^2011-3^2012=-3
có:2012 số hạng=> 2012:2=1006(cặp)
Vậy S2=1006.(-3)=-3018
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s1=1+2+3+...+99
s1=99+98+...+1
2s1=100+100+....+100
2s1=100.99
s1=100.99:2=4950(mấy bài sau lam tương tự nha)
4+4^2+4^3+...+4^90 chia hết cho 21
=(4+4^2+4^3)+...+(4^88+4^89+4^90)
=84.1+(4^4+4^5+4^6+...+4^90)
vì 84 chia hết cho 21 suy ra tổng trên chia hét cho 21 (ĐPCM)
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a, 144 =(142)2 =1962 ; 1=12
=> 144 - 1 =1962 - 12 =(196 -1)2
=1952 Mà 1952 chia hết cho 3 nên => 144 - 1 chia hết cho 3
b, Ta có :
A= 2+22+23+.....+260
A=(2+22+23+24)+(25+26+27+28)+.....+(256+257+258+259260)
A=2(1+22+23)+25(1+22+23)+.....+256(1+22+22+23)
A=2*15+25*15+.....+256*15
A=15(2+25+.....+256) chia hết cho 15
nhớ **** cho mk nka !
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\(S=1+2+2^2+...+2^{99}\)
\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)
\(S=3+2^2.3+...+2^{98}.3\)
\(=3\left(1+2^2+...+2^{98}\right)⋮3\)