Tính 1/3 +1/3^2+1/3^3+.....+1/3^50
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
E=-1/3+1/3^2-1/3^3+1/3^4-...+1/3^50-1/3^51
3E=-1+1^2-1^3+1^4-1^5+...+1^50-1^51
3E=-1+1-1+1-1+...+1-1
3E=0
![](https://rs.olm.vn/images/avt/0.png?1311)
-----A=-1/3+1/3^2-1/3^3+-----+1/3^50-1/...
A*1/3 =-1/3^2+1/3^3+--------------------+1/3^5...
--------A=-1/3+1/3^2-1/3^3+...+1/3^50-1...
-------A*1/3 =-1/3^2+1/3^3+..---------...+1/3^51-1/3^...
---------------------------------------...
A+A*1/3=-1/3+0...+0+...0---------------...
A+A*1/3= -1/3-1/3^52
4/3*A= -1/3-1/3^52
Vậy
A= -(1/3+1/3^52)*3/4.
![](https://rs.olm.vn/images/avt/0.png?1311)
B=-1/3+1/3^2-.....-1/3^51
3B=-1/3^2+1/3^3-.....-1/3^52
3B-B=(-1/3^2+1/3^3-....-1/3^52)-(-1/3+1/3^2-....-1/3^51)
2B= -1/3^52-1/3
2B= -1/3^52-3^51/3^52
2B= -1-3^51/3^52
B= -3^51-1/3^52x2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{50}}-\dfrac{1}{3^{51}}\)
\(=\dfrac{1}{\left(-3\right)}+\dfrac{1}{\left(-3\right)^2}+\dfrac{1}{\left(-3\right)^3}+...+\dfrac{1}{\left(-3\right)^{50}}+\dfrac{1}{\left(-3\right)^{51}}-\dfrac{1}{3}\)
\(=\dfrac{1}{\left(3\right)^2}+\dfrac{1}{\left(3\right)^3}+...+\dfrac{1}{\left(-3\right)^{51}}+\dfrac{1}{\left(-3\right)^{52}}\)
\(\Rightarrow\dfrac{4}{3}B=\dfrac{1}{-3}-\dfrac{1}{\left(-3\right)^{52}}=\dfrac{-3^{51}-1}{3^{52}}\Rightarrow B=\dfrac{-3^{51}-1}{4.3^{51}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1 × 2 × 3 + 2 × 3 × 4 + .....+ 48 × 49 × 50
ta có 4 x A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 -1) + .....+ 48 × 49 × 50 x (51 - 47)
= 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + ... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50
= 48 x 49 x 50 x 51
suy ra A = (48 x 49 x 50 x 51) : 4
= 12 x 49 x 50 x 51
nhớ k cho mik nha rùi mik lm nốt cho
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{50}}\)
=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{49}}\)
=>\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{49}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{50}}\right)\)
=>2A=\(1-\frac{1}{3^{50}}\)
=>A=\(\frac{1-\frac{1}{3^{50}}}{2}\)
\(=>A=\frac{1}{2}-\frac{1}{\frac{3^{50}}{2}}=\frac{1}{2}-1.\frac{2}{3^{50}}=\frac{1}{2}-\frac{2}{3^{50}}=\frac{3^{50}-4}{2.3^{50}}\)
Vậy..................
hello