A = 22 + 23 + 24 + ... + 2121
⇒ A = (22 + 23) + (24 + 25) + ... + (2120 + 2121)
⇒ A = 12 + 22(22 + 23) +... + 2118(22 + 23)
⇒ A = 12 + 22.12 + ... + 2118.12
⇒ A = 12(1 + 22 + ... + 2118) ⋮ 3
⇒ A ⋮ 3
Vì: A co 120 so hạng nên ta chia A thành 60 nhoms moi nhoms co 2 so hạng như sau:
\(A=2^2+2^3+\:2^4+\:2^5+\:..+\:2^{121}=2^2\left(1+2\right)+2^4\left(1+2\right)+......+2^{120}\left(1+2\right)=2^2.3+2^4.3+......+2^{120}.3=3\left(2^2+2^4+....+2^{120}\right)⋮3\Rightarrow A⋮3\)
Vì: A co 120 so hạng nên ta chia A thành 40 nhoms moi nhoms co 3 so hạng như sau:
\(A=\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+....+\left(2^{119}+2^{120}+2^{121}\right)=2^2\left(1+2+4\right)+2^5\left(1+2+4\right)+.....+2^{119}\left(1+2+4\right)=2^2.7+2^5.7+...+2^{119}.7=7\left(2^2+2^5+....+2^{119}\right)⋮7\Rightarrow A⋮7\)
đáp án nek:
Ta có: A= 22+23+24+...+2121
= (22+23)+(24+25)+...+(2120+2121)
22.(1+2)+24.(1+2)+...+2120.(1+2)
22.3+24.3+...+2120.3
3.(22+24+...+2120) chia hết cho 3
=> A chia hết cho 3
Có: A= 22+23+24+...+2121
= (22+23+24)+(25+26+27)+...+(2199+2120 +2121)
= 22.(1+2+4)+25.(1+2+4)+...+2199.(1+2+4)
=b22.7+25.7+...+2199.7
= 7.(22+25+...+2199) chia hết cho 7
=> A chia hết cho 7
A = 22 + 23 + 24 + ... + 2121
A = (22 + 23) + (24 + 25) + ... + (2120 + 2121)
A = 1 . (22 + 23) + 22 . (22 + 23) + ... + 2118 . (22 + 23)
A = (1 . 12) + (22 . 12) + ... + (2118 . 12)
A = 12 . (1 + 22 + ... + 2118) ⋮ 3
=> A ⋮ 3
A = 22 + 23 + 24 + ... + 2121
A = (22 + 23 + 24) + (25 + 26 + 27) + ... + (2119 + 2120 + 2121)
A = 22 . (1 + 2 + 4) + 25 . (1 + 2 + 4) + ... + 2119 . (1 + 2 + 4)
A = (22 . 7) + (25 . 7) + ... + (2119 . 7)
A = 7 . (22 + 25 + ... + 2119) ⋮ 7
=> A ⋮ 7
