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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
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
\(S=3+3^2+3^3+...+3^{2019}\)
\(3S=3^2+3^3+...+3^{2019}+3^{2020}\)
\(\Rightarrow3S-S=-3+3^{2020}\)
\(\Rightarrow2S=3^{2020}-3\Rightarrow S=\frac{3^{2020}-3}{2}\)
Ta có: \(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2017}+3^{2018}+3^{2019}\right)\)
\(=3\left(1+3+9\right)+3^4\left(1+3+9\right)+...+3^{2017}\left(1+3+9\right)\)
\(=3.13+3^4.13+...+3^{2017}.13\)
\(=13.\left(3+3^4+...+3^{2017}\right)⋮13\)
bạn gom 3 số 1 lần là giải dc bài thôi
\(S=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^7\left(1+3+3^2\right)+3^{10}\left(1+3+3^2\right)+3^{13}\left(1+3+3^2\right)\)=\(\left(1+3+3^2\right)\left(3+3^4+3^7+3^{10}+3^{13}\right)=13A\)chia het cho 13