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A= \(2^0+2^1+2^2+...+2^{50}\)
\(\Rightarrow\)2A =2(\(2^0+2^1+2^2+...+2^{50}\))
\(\Rightarrow\)2A= \(2+2^2+2^3+2^4+...+2^{51}\)
\(\Rightarrow\)2A-A= (\(2+2^2+2^3+2^4+...+2^{51}\))-(\(2+2^2+2^3+2^4+...+2^{50}\))
\(\Rightarrow\)A= \(2^{51}-1\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{15}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{15}}\right)\)
\(A=1-\frac{1}{2^{15}}\)\
\(A=1-\frac{1}{32768}\)
\(A=\frac{32767}{32768}\)
a, S = 2 + 22 + 23 + 24 + ... + 299 + 2100. 2S = 22 + 23 + 24 + 25 + ... + 2100 + 2101 => 2S - S = S = (22 + 23 + 24 + 25 + ... + 2100 + 2101) - (2 + 22 + 23 + 24 + ... + 299 + 2100) = 2101 - 2. Vậy S = 2101 - 2. b, S = 2 + 22 + 23 + 24 + ... + 299 + 2100 = (2 + 22) + (23 + 24) + ... + (299 + 2100) = 2.(1 + 2) + 23.(1 + 2) + ... + 299.(1 + 2) = (1 + 2).(2 + 23 + ... + 299) = 3.(2 + 23 + ... + 299) => S ⋮ 3. Vậy S ⋮ 3 (đpcm)
\(2^{x+1}+5.2^{x+2}=88\)
\(\Rightarrow2^x.2+5.2^x.2^2=88\)
\(\Rightarrow2^x\left(2+5.2^2\right)=88\)
\(\Rightarrow2^x.22=88\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy x = 2
Theo bài ra ta có:
2x+1+5.2x+2 = 88
=> 2x.2+5.2x.22
=> 2x.( 2 = 5.22 )
=> 2x.22 = 88
=> 2x = 88:22
=> 2x = 4
=> x = 4:2
=> x = 2
Vậy x = 2