cho a=2+2^2+2^3 +........+2^90 chứng minh A chia hết cho 14
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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
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
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a) Ta có:
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=14+...+2^{21}.\left(2+2^2+2^3\right)\)
\(\Rightarrow A=14+...+2^{21}.14\)
\(\Rightarrow A=\left(1+...+2^{21}\right).14⋮14\)( đpcm )
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{21}+2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{21}\left(1+2+2^2+2^3\right)\)
\(\Rightarrow A=2.15+...+2^{21}.15\)
\(\Rightarrow A=15\left(2+...+2^{21}\right)⋮15\left(đpcm\right)\)
b) Mk sửa đề chút là A chia 16 dư 15 nhé
Ta có:
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{20}+2^{21}+2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{20}\left(1+2+2^2+2^3+2^4\right)\)
\(\Rightarrow A=2.31+...+2^{20}.31\)
\(\Rightarrow A=\left(2+2^{20}\right).31\)
Vì 31 chia 16 dư 15 nên suy ra đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+...+2^{2019}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+^9\right)+...+\left(2^{2017}+2^{2018}+2^{2019}\right)\)
\(=14+2^4\left(2+2^2+2^3\right)+2^7\left(2+2^2+2^3\right)+...+2^{2017}\left(2+2^2+2^3\right)\)
\(=14+2^4.14+2^7.14+...+2^{2017}.14\)
\(=14\left(1+2^4+2^7+...+2^{2017}\right)⋮14\)
\(\Rightarrow A⋮14\)
#_ARMY_#
![](https://rs.olm.vn/images/avt/0.png?1311)
A=21+22+23+...+261+262+263
A=(21+22+23)+...+(261+262+263)
A=14+...+261.(21+22+23)
A=14+...+261.14 chia hết cho 14
tick ủng hộ mình nha