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A = 3+32+33+.....+3100
3A = 32+33+34+....+3101
2A = 3A - A = 3101-3 < 3101
=> A = \(\frac{3^{101}-3}{2}<3^{101}\)
=> A < B
A = 3 + 32 + 33 + 34 +.............3100
3A =32 + 33 + 34 +.............3101
3A - A = (3 + 32 + 33 + 34 +.............3100) - (32 + 33 + 34 +.............3101)
2A = 3101 - 3
\(A=\frac{3^{101}-3}{2}\)
B = 3101
Ta có A < B
\(3^{21}=3.3^{20}=3.\left(3^2\right)^{10}=3.9^{10}\)
\(2^{31}=2.2^{30}=2.\left(2^3\right)^{10}=2.8^{10}\)
Thấy: 3 > 2 và 910 > 810
Nên \(3^{21}>2^{31}\)
Bài 2:
\(A=1+2+2^2+.....+2^{100}\)
\(2A=2+2^2+.......+2^{101}\)
\(2A-A=\left(2-2\right)+\left(2^2-2^2\right)+......+2^{101}-1\)
Vậy A = 2101 - 1
a)A=3^0+3^1+3^2+3^3+...+3^2012
A=1+3+3^2+3^3+..+3^2012
3A=3+3^2+3^3+3^4+..+3^2013
3A-A=3+3^2+3^3+3^4+..+3^2013-1-3-3^2-3^3-...-3^2012
2A=3^2013-1
A=\(\frac{3^{2013}-1}{2}\)
B=3^2013
=> A>B
b) A=1+5+5^2+5^3+..+5^99+5^100
5A=5+5^2+5^3+5^4+...+5^100+5^101
5A-A=5+5^2+5^3+5^4+..+5^100+5^101-1-5-5^2-5^3-..-5^99-5^100
4A=5^101-1
A=\(\frac{5^{101}-1}{4}\)
B=5^101/4
=> A<B
\(3+3^2+.......+3^{99}\)
\(\Rightarrow3\left(3+3^2+.....+3^{99}\right)=3^2+3^3+......+3^{100}\)
\(\Rightarrow\left(3^2+3^3+.....+3^{100}\right)-\left(3+3^2+....+3^{99}\right)\)
\(=3^{100}-3\)
\(Do3^{100}-3< 3^{100}=>3+3^2+....+3^{99}< 3^{100}\)
Đặt
\(A=3+3^2+...+3^{99}\)
\(B=3^{100}\)
\(3A=3^2+3^3+...+3^{100}\)
\(3A-A=\left(3^2+3^3+...+3^{100}\right)-\left(3+3^2+...+3^{99}\right)\)
\(2A=3^{100}-3\)
\(A=\frac{3^{100}-3}{2}\) mà \(B=3^{100}\)
\(\Rightarrow A< B\)