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ta có :
2.S=\(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\)
2.S-S=\(\left(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
=>S=\(\frac{3}{2^{10}}-\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: S = 3+3/2+3/2^2+3/2^3+...+3/2^9
1/2.S = 3/2+3/2^2+3/2^3+3/2^4+...+3/2^10
\(\Rightarrow\) S-1/2.S = 3 - 3/2^10
\(\Rightarrow\) 1/2.S = 3 - 3/2^10
\(\Rightarrow\) S = (3 - 3/2^10) : 1/2
\(\Rightarrow\) S = 6 - 6/2^10
Nếu đúng thì cho mk biết nha
![](https://rs.olm.vn/images/avt/0.png?1311)
2S=32+33+34+....+32016
2S-S=(32+33+34+...+32016)-(3+32+33+....+32015)
S=22016-3
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Lời giải chi tiết:
Có: S=3+32+322+323+...+329.
2S=6+3+32+322+....+328.
Trừ vế với vế của hai biểu thức trên và triệt tiêu các hạng tử giống nhau, ta được:
2S−S=6−329=6−3512=3069512, suy ra S=3069512.
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\(\Rightarrow2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(\Rightarrow2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)\)
\(\Rightarrow S=6-\frac{3}{2^9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(=3\left(2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2^9}\right)=3\left(2-\frac{1}{2^9}\right)=6-\frac{3}{2^9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=3\left(1+\frac{1}{2^{ }}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(2S=3\left(\frac{2}{2^0}+\frac{2}{2^1}+\frac{2}{2^2}+...+\frac{2}{2^9}\right)=3\left(2+1+\frac{1}{2^{ }}+...+\frac{1}{2^8}\right)\)\(2S-S=S=3\left(2+1+\frac{1}{2^1}+...+\frac{1}{2^8}\right)-3\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=3.\left(2-\frac{1}{2^9}\right)=3.\frac{2^{10}-1}{2^9}\)