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B = 2 + 22 + 23 + ... + 210
=> 2B = 22 + 23 + 24+ ... + 211
Lấy 2B trừ B theo vế ta có :
2B - B = ( 22 + 23 + 24+ ... + 211) - (2 + 22 + 23 + ... + 210)
B = 211 - 2
B + 2 = 211 - 2 + 2
= 211
Vậy B + 2 = 211 (đpcm)
\(2B=2^2+2^3+.....+2^{11}\)
\(B=2^{11}-2\)
\(B+2=2^{11}⋮2^{11}\)
\(\Rightarrow B+2⋮2^{11}\left(đpcm\right)\)
b)
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}\left(4.13+4.65\right)}{2^{10}.104}+\frac{3^9\left(11.3+5.3\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}=3+3=6\)
a) \(A=4+2^2+2^3+2^4+....+2^{20}\)
\(\Rightarrow2A=2^3+2^3+2^4+.....+2^{21}\)
\(\Rightarrow2A-A=\left(2^3+2^3+2^4+....+2^{21}\right)-\left(2^2+2^3+2^4+...+2^{20}\right)\)
\(\Rightarrow A=2^3+2^{21}-\left(2^2+2^2\right)\)
\(\Rightarrow A=2^{21}\)
\(\text{Vì }2^{21}⋮2^7\Rightarrow A⋮128\)
b) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{12}\left(13+65\right)}{2^{10}.2^3.13}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)
\(=\frac{2^{12}.78}{2^{13}.13}+\frac{3^{10}.16}{3^9.16}=\frac{6}{2}+\frac{3^{10}}{3^9}\)
\(=3+3=6\)