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a) Ta có: \(A=1+3+3^2+...+3^{99}+3^{100}\)
=> \(3A=3+3^2+3^3+...+3^{100}+3^{101}\)
=> \(3A-A=\left(3+3^2+...+3^{101}\right)-\left(1+3+...+3^{100}\right)\)
<=> \(2A=3^{101}-1\)
=> \(A=\frac{3^{101}-1}{2}\)
b) Ta có: \(B=1+4+4^2+...+4^{100}\)
=> \(4B=4+4^2+4^3+...+4^{101}\)
=> \(4B-B=\left(4+4^2+...+4^{101}\right)-\left(1+4+...+4^{100}\right)\)
<=> \(3B=4^{101}-1\)
=> \(B=\frac{4^{101}-1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt A = 21 + 22 + 23 +....+299 + 2100
=> 2A = 2+ 22 + 23 +...+2100 + 2101
=> 2A - A = 2 + 22 + 23 +...+2100 + 2101 - (1+2+22+23+...+299+2100)
=> A = 2 + 22 + 23 +...+ 2100 +2101 -1 - 2 - 22 - 23 -...- 299 - 2100
= 2101 -1
Vậy....
b) B = 2 + 23 + 25 + ... + 22013
4B = 23 + 25 + 27 + ... + 22015
4B - B = (23 + 25 + 27 + ... + 22015) - (2 + 23 + 25 + ... + 22013)
3B = 22015 - 2
B = \(\dfrac{2^{2015}-2}{3}\)
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A=1+3+32+33+34+35+………..+399+3100
3A = 3+32+33+34+35+………..+3100+3101
3A - A = ( 3+32+33+34+35+………..+3100+3101 ) - ( 1+3+32+33+34+35+………..+399+3100 )
2A = 3101 - 1
A = \(\frac{3^{101}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+3^1+3^2+3^3+3^4+....+3^{99}+3^{100}\)
\(\Rightarrow3A=3+3^2+2^3+3^4+....+3^{100}+3^{101}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+....+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-1\)
\(\Rightarrow A=\frac{3^{101}-1}{2}\)
\(A=1+3^1+3^2+...+3^{99}+3^{100}\)
\(3A=3+3^1+3^2+...+3^{99}+3^{100}\)
\(3A-A=\left(3+3^2+3^3+...+3^{100}+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\left(3^{101}-1\right):2\)