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bai 3
\(A=\frac{10^{2004}+1}{10^{2005}+1}\)
\(10A=\frac{10^{2004}+10}{10^{2005}+1}\)
\(10A=1\frac{9}{10^{2005}+1}\)
\(B=\frac{10^{2005}+1}{10^{2006}+1}\)
\(10B=\frac{10^{2005}+10}{10^{2006}+1}\)
\(10B=1\frac{9}{10^{2006}+1}\)
Vì \(1\frac{9}{10^{2005}+1}>1\frac{9}{10^{2006}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
bai 4
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^8}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^9}\)
\(A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^9}\)
A = 1 + 3 + 32 + ... + 310
3A = 3 + 32 + ... + 311
3A - A = 311 - 1
2A = 311 - 1
Thay vào đề :
311 - 1 + 1 = 3n
311 = 3n
=> n = 11
3A=3(1+3+32+.....+310)
3A=3+32+33+34+....+311
3A-A=(3+32+33+34+....+311)-(1+3+32+.....+310)
2A=311-1
=>2A+1=311-1+1=311
Vậy n=11
\(A=1+3+3^2+3^3+...+3^{10}\)
\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{11}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{11}\right)-\left(1+3+3^2+3^3+...+3^{10}\right)\)
\(\Rightarrow2A=3+3^2+3^3+...+3^{11}-1-3-3^2-3^3-...-3^{10}\)
\(\Rightarrow2A=3^{11}-1\)
\(\Rightarrow2A+1=3^{11}-1+1=3^{11}\) (1)
mà : \(2A+1=3^n\) (2)
Từ (1) và (2) \(\Rightarrow3^{11}=3^n\Rightarrow n=11\)
Vậy : \(n=11\) khi \(2A+1=3^n\)
a=1+3+32+33+...+310
3a=3.(1+3+32+33+...+310)
3a=3+32+33+34+...+311
3a-a=(3+32+33+34+...+311)-(1+3+32+33+...+310)
2a=311-1
a=(311-1):2
a=88573
2.88573+1=3n
177146+1=3n
311=3n
=>n=11