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\(A=2^1+2^2+2^3+...+2^{2010}\)
\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+...+2^{2010}\)
\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)
A=21+22+23+...+22010
=(21+22)+(23+24)+...+(22009+22010)=(21+22)+(23+24)+...+(22009+22010)
=2(1+2)+23(1+2)+...+22009(1+2)=2(1+2)+23(1+2)+...+22009(1+2)
=3(2+23+...+22009)⋮3=3(2+23+...+22009)⋮3
�=21+22+23+...+22010A=21+22+23+...+22010
=(21+22+23)+(24+25+26)+...+(22008+22009+22010)=(21+22+23)+(24+25+26)+...+(22008+22009+22010)
=2(1+2+22)+24(1+2+22)+...+22008(1+2+22)=2(1+2+22)+24(1+2+22)+...+22008(1+2+22)
=7(2+24+...+22008)⋮7=7(2+24+...+22008)⋮7
Lời giải:
$A=1+3+3^2+3^3+....+3^{2026}$
$=1+3+3^2+(3^3+3^4+3^5+3^6)+(3^7+3^8+3^9+3^{10})+....+(3^{2023}+3^{2024}+3^{2025}+3^{2026})$
$=13+3^2(3+3^2+3^3+3^4)+3^6(3+3^2+3^3+3^4)+...+3^{2022}(3+3^2+3^3+3^4)$
$=13+(3^2+3^6+...+3^{2022})(3+3^2+3^3+3^4)$
$=13+(3^2+3^6+...+3^{2022}).120$
$\Rightarrow A$ chia $120$ dư $13$
sao ko dung f(x) ma viet
\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^9+2^{10}\)
a=\(\left(2+2^2\right)+2^2.\left(2+2^2\right)+..+2^8\left(2+2^2\right)\)
a=\(\left(2+2^2\right).\left(1+2^2+..+2^8\right)\)
a=\(6.\left(1+2^2+2^4+2^6+2^8\right)\)
chia het cho 3
\(10^6\) tận cùng là 0 \(=>10^6+2\) tận cùng là 2 \(=>10^6+2\) chia hết cho 2
A=3+32+33+...+320
3A=3.(3+32+33+...+320)
3A=32+33+34+...+321
3A-A=(32+33+34+...+321)-(3+32+33+...+320)
2A=321-3
A=\(\frac{3^{21}-3}{2}\)
B=\(\frac{3^{21}}{2}\)
=>B-A=\(\frac{3^{21}}{2}\)-\(\frac{3^{21}-3}{2}\)=\(\frac{3}{2}\)=1,5
Chúc bn học tốt
Có 3A = 3 mũ 2 + 3 mũ 3 +...+ 3 mũ 20 + 3 mũ 21
- A = 3 + 3 mũ 2 + 3 mũ 3 +... + 3 mũ 20
2A = 3 mũ 21 - 3
A = (3 mũ 21 - 3) : 2
\(\Rightarrow\)B-A=(3 mũ 21 : 2) - [(3 mũ 21 - 3):2]
B-A=3 mũ 21 - (3 mũ 21 - 3)
B-A=3 mũ 21- 3 mũ 21 +3
B-A=3
Vậy B-A=3
Đây là ý kiến của mik thôi! Mik ko bt là mik có tính sai ko! Mong mọi người góp ý!
Chúc bn hok tốt!
tìm số dư của
A= [22^6n+2(hai mũ hai mũ sáu n cộng hai) + 3]:7
B =[22^3n+1(hai mũ ba n cộng một)+3]:13
\(3+3^2+3^3+...+3^{60}\\ =\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\\ =\left(1+3\right)\left(3+3^3+...+3^{59}\right)\\ =4\left(3+3^3+...+3^{59}\right)⋮4\\ 3+3^2+3^3+...+3^{60}\\ =\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ =3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\left(3+3^4+...+3^{58}\right)\\ =13\left(3+3^4+...+3^{58}\right)⋮13\)
Lời giải:
$A=1+3+3^2+(3^3+3^4+3^5+3^6)+(3^7+3^8+3^9+3^{10})+...+(3^{2023}+3^{2024}+3^{2025}+3^{2026})$
$=13+3^2(3+3^2+3^3+3^4)+3^6(3+3^2+3^3+3^4)+...+3^{2022}(3+3^2+3^3+3^4)$
$=13+(3+3^2+3^3+3^4)(3^2+3^6+...+3^{2022})$
$=13+120(3^2+3^6+...+3^{2022})$
Suy ra $A$ chia $120$ dư $13$