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\(A=3+3^2+3^3+...+3^{120}\)
\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{100}\right)\)
\(3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^3+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}-3+3=3^{101}=3^n\)
\(\Rightarrow n=101\)
vậy ...
\(3A=3^2+3^3+...+3^{121}\)
\(3A-A=\left(3^2-3^2\right)+........+\left(3^{120}-3^{120}\right)+3^{121}-3\)
A = \(\frac{3^{121}-3}{2}\)
2A + 3 = \(\frac{3^{121}-3}{2}.2+3=3^{121}=3^n\)
Vậy n = 121
\(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2013}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\)
\(\Rightarrow3B=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\right)\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2012}}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2012}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^{2013}}\Rightarrow1-2B=\frac{1}{3^{2013}}=\left(\frac{1}{3}\right)^{2013}\Rightarrow n=2013\)
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013