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\(A=1+4+4^2+...+4^{59}\)
\(A=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{58}+4^{59}\right)\)
\(A=\left(1+4\right)+4^2\left(1+4\right)+...+4^{58}\left(1+4\right)\)
\(A=5+4^2.5+...+4^{48}.5\)
\(A=5\left(1+4^2+...+4^{48}\right)\)
\(\Rightarrow A⋮5\)
\(A=1+4+4^2+...+4^{59}\)
\(A=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{57}+4^{58}+4^{59}\right)\)
\(A=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{47}\left(1+4+4^2\right)\)
\(A=21+4^3.21+...+4^{47}.21\)
\(A=21\left(1+4^3+...+4^{47}\right)\)
\(\Rightarrow A⋮21\)
\(A=\left(3^{60}+3^{58}+3^{56}+...+3^2\right)-\left(3^{59}+3^{57}+3^{55}+...+3\right).\)
\(B=3^{60}+3^{58}+3^{56}+...+3^2\)
\(9B=3^{62}+3^{60}+3^{58}+...+3^4\)
\(B=\frac{9B-B}{8}=\frac{3^{62}-3^2}{8}=\frac{3^2\left(3^{60}-1\right)}{8}\)
\(C=3^{59}+3^{57}+3^{55}+...+3\)
\(9C=3^{61}+3^{59}+3^{57}+...+3^3\)
\(C=\frac{9C-C}{8}=\frac{3^{61}-3}{8}=\frac{3\left(3^{60}-1\right)}{8}\)
\(A=B-C=\frac{3^2\left(3^{60}-1\right)-3\left(3^{60}-1\right)}{8}=\frac{6\left(3^{60}-1\right)}{8}\)
\(A=\frac{2.3.\left(3^{60}-1\right)}{8}=\frac{2.3.3^{60}}{8}-\frac{2.3}{8}=\frac{3^{61}}{4}-\frac{3}{4}=\frac{3^{61}-3}{4}\)
Mẫu câu a)!! những câu khác ko lm đc ib!
a) Ta có:
\(A=2+2^2+2^3+2^4+...+2^{2010}.\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{2009}.3\)
\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)
Ta có:
\(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{2008}.7\)
\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)
b,\(B=3+3^2+3^3+3^4+...+3^{2010}.\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{2009}.4\)
\(=4.\left(3+3^3+...+3^{2009}\right)⋮4\)
\(B=3+3^2+3^3+3^4+...+3^{2010}\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{2008}.13\)
\(=13\left(3+3^4+...+3^{2008}\right)⋮13\)
a.A= 3+ 32+ 33 + 34 +...+310
Ta có :A= 3 + 32 + 33 + 34 + ... +310
A= 3+ 9+ 27+ 81+ ...+310
A= (3 +9)+(33 + 34)+(35 + 36)+...+(39 + 310)
A= 12 + (32 X 3 +32 X 32) + (34 X 3 + 34 X 32) + ...+ (38 X 3 + 38 X 32)
A= 12 + [32 X (3 + 32)] + [34 X (3+32)] + ....+ [38X(3 + 32)]
A= 12 + 32 X 12 + 34 X 12 + .... + 38 X 12
A= 12 X (1 + 32 + 34 + ... + 38)
Vì 12 chia hết cho 4 nên A chia hết cho 4
\(1+4+4^2+4^3+...+4^{58}+4^{59}\)
\(=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{58}+4^{59}\right)\)
\(=5+\left(4^2.1+4^2.4\right)+....+\left(4^{58}.1+4^{58}.4\right)\)
\(=5+4^2.\left(1+4\right)+...+4^{58}.\left(1+4\right)\)
\(=1.5+4^2.5+....+4^{58}.5\)
\(=\left(1+4^2+...+4^{58}\right).5⋮5\)
LÀM ƠN NHANH LÊN MIK VỘI LẮM
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