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B = (1 + 3) + (32+33)+.....+(389+390)
= 4 + 32 .(1 + 3) + .....+390.(1+3)
= 1 .4 + 32.4 + ..... +390.4
= 4.(1 + 32 + .... +390) chia hết cho 4
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
bạn nhóm 3 số vào 1 nhóm rồi nhóm ts chung riêng nhóm thứ nhất tính ra lun
Giải
Ta có: S=\(3^0+3^2+3^4+...+3^{2002}\)
\(\Leftrightarrow\)\(3^2\)S=\(3^2\)(\(3^0+3^2+3^4+...+3^{2002}\))
\(\Leftrightarrow\)\(3^2S=3^2+3^4+3^6+...+3^{2004}\)
\(\)\(3^2S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+3^4+...+3^{2002}\right)\)
8S=\(\left(3^2-3^2\right)+\left(3^4-3^4\right)+\left(3^6-3^6\right)+...+\left(3^{2002}-3^{2002}\right)+3^{2004}-1\)
8S=0+0+0+...+\(3^{2004}\)-1=\(3^{2004}-1\)
\(\Leftrightarrow\)S=\(\frac{3^{2004}-1}{8}\)
\(S=3+3^2+3^3+...+3^{2019}\)
\(3S=3^2+3^3+...+3^{2019}+3^{2020}\)
\(\Rightarrow3S-S=-3+3^{2020}\)
\(\Rightarrow2S=3^{2020}-3\Rightarrow S=\frac{3^{2020}-3}{2}\)
Ta có: \(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2017}+3^{2018}+3^{2019}\right)\)
\(=3\left(1+3+9\right)+3^4\left(1+3+9\right)+...+3^{2017}\left(1+3+9\right)\)
\(=3.13+3^4.13+...+3^{2017}.13\)
\(=13.\left(3+3^4+...+3^{2017}\right)⋮13\)
S=1+2+2^2+2^3+...+2^59
S=(1+2)+(2^2+2^3)+...+(2^58+2^59)
S=3+2^2(1+2)+...+2^58.(1+2)
S=3+2^2.3+...+2^58.3
S= 3.( 1+2^2+...+2^58) chia hết cho 3
S=1+2+2^2+2^3+...+2^59
S=(1+2+2^2)+(2^3+2^4+2^5)+...+(2^57+2^58+2^59)
S=7.2^3(1+2+2^2)+....+2^57(1+2+2^2)
S=7+2^3.7+...+2^57.7
S=7.(1+2^3+...+2^57) chia hết cho 7
S= 1+2+2^2+2^3+...+2^59
S=(1+2+2^2+2^3)+(2^4+2^5+2^6+2^7)+...+(2^56+2^57+2^58+2^59)
S=15+2^4(1+2+2^2+2^3)+...+2^56(1+2+2^2+2^3)
S=15+2^4.15+...+2^56.15
S=15(1+2^4+...+2^56) chia hết cho 15
chắc chắn đúng tick cho mình nhé!
S = 1 + 3 + 32 + 33 + ... + 359
S = ( 1 + 3 + 32 ) + ( 33 + 34 + 35 ) + ... + ( 357 + 358 + 359 )
S = 13 + 33( 1 + 3 + 32 ) + ... + 357( 1 + 3 + 32 )
S = 13 + 33 . 13 + ... + 357 . 13
S = 13 ( 1 + 33 + ... + 357 ) ⋮ 13 vì 13 ⋮ 13
Vậy S ⋮ 13