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\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
\(A=\frac{2^{100}-1}{2^{100}}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^n}\)
\(3S=3+1+\frac{1}{3}+.......+\frac{1}{3^{n-1}}\)
\(\Rightarrow3S-S=\left(3+1+\frac{1}{3}+......+\frac{1}{3^{n-1}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^n}\right)\)
\(\Rightarrow2S=3-\frac{1}{3^n}\Rightarrow2S=\frac{3^{n+1}-1}{3^n}\Rightarrow S=\frac{3^{n+1}-1}{2.3^n}\)
a, A = 3/2 × 4/3 × 5/4 × ... × 81/80
A = 81/2
b) (1 - 1/2) × (1 - 1/3) × ... × (1 - 1/100)
= 1/2 × 2/3 × .. × 99/100
= 1/100
C = (-1/2).(-2/3). ...... .[-(n-1)/n]
+, Nếu n lẻ thì :
C = 1/2.2/3. .... . (n-1)/n = 1/n
+, Nếu n chẵn thì :
C = -[1/2.2/3. ..... . (n-1)/n] = -1/n
Vậy .............
Tk mk nha
\(\Rightarrow C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n-1}{n}\)
\(\Rightarrow C=\frac{1}{n}\)
1,
Tỉ số giữa 10 quyển và 15 quyển:
10: 15 = 2/3
Nếu chia đều thì mỗi bạn nhận đc:
[15x 2 + 10x3] : [2+3] = 12 [quyển]
Vậy:....................
2,
1/2 + 1/3 + 1/4 + ... + 1/50 = [1 - 1/2] + [1-2/3] + ... + [1 - 49/50]
= 1 - 1/2 + 1 - 2/3 + ... + 1 - 49/50
= [1 + 1 + 1 +... + 1] - [1/2+2/3+3/4+...+49/50]
= 49 - [1/2+2/3+3/4+...+49/50]
Vậy 1/2 + 1/3 + 1/4 + ... + 1/50 không là số tự nhiên
3,
1/42 + 1/52 + ... +1/1002 < 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/99.100
<=> 1/42 + 1/52 + ... +1/1002 < 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
<=> 1/42 + 1/52 + ... +1/1002 < 1/3 - 1/100
<=> E < 1/3 - 1/100
=> E < 1/3
Mà 1/3 - 1/100 = 97/300 > 1/5
=> 1/5 < E < 1/3
4, A:
2013/1 + 2014/2+2015/3+...+4023/2011+4024/2012 - 2012
= ( 2013/1 - 1)+(2014/2 - 1) + ( 2015/3 - 1)+...+ (4023/2011 - 1) + ( 4024/2012 - 1)
= 2012(1+1/2+1/3+...+ 1/2011+1/2012)
Vậy \(A=\frac{\text{(1+1/2+1/3+...+ 1/2011+1/2012)}}{\text{2012(1+1/2+1/3+...+ 1/2011+1/2012)}}=\frac{1}{2012}\)
Câu B mik sẽ làm sau, bây giờ mik bận
Tỉ số giữa 10 quyển và 15 quyển:
10:15=2/3
Vậy nếu chia cho cả lớp thì mõi bạn nhận được:
(15x2+10x3):5=12 quyển
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{2011}}\)
\(\Rightarrow2A-A=2-\frac{1}{2^{2012}}\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
Ta có: A = \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
2A = \(2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
2A = \(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
2A - A = \(\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
A = \(2-\frac{1}{2^{2012}}\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
- Bn lấy 2A - A = A là ra nhé :))
Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}+\frac{1}{2^{2012}}\)
\(\Rightarrow2A-A=A=2-\frac{1}{2^{2012}}\)