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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: \(B=1+4+4^2+...+4^{100}\)
=> \(4B=4+4^2+4^3+...+4^{101}\)
=> \(4B-B=\left(4+4^2+...+4^{101}\right)-\left(1+4+...+4^{100}\right)\)
<=> \(3B=4^{101}-1\)
=> \(B=\frac{4^{101}-1}{3}\)
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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 + 2 + 22 + 23 + ... + 299 + 2100 . (1)
\(\Rightarrow2A=2+2^2+2^3+...+2^{101}\) (2)
Trừ 2 vế của (1) và (2) cho nhau được \(A=2^{101}-1\)
Ta có: A = 1 + 2 + 22 + 23 + ... + 299 + 2100.
2A = 2 (1 + 2 + 22 + 23 + ... + 299 + 2100)
= \(2\cdot1+2\cdot2+2\cdot2^2+2\cdot2^3+...+2\cdot2^{99}+2\cdot2^{100}.\)
2A = \(2+2^2+2^3+2^4+...+2^{100}+2^{101}.\)
2A - A = \(\left(2+2^2+2^3+2^4+...+2^{100}+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)
A = \(2^{101}-1\).
Vậy A = 2101 - 1.
\(M=1+2+2^2+2^3+2^4+...+2^{99}\)
\(\Rightarrow2M=2\left(1+2+2^2+2^3+2^4+...+2^{99}\right)\)
\(\Rightarrow2M=2+2^2+2^3+2^4+2^5+...+2^{100}\)
\(\Rightarrow2M-M=\left(2+2^2+2^3+2^4+2^5+...+2^{100}\right)-\left(1+2+2^2+2^3+2^4+...+2^{99}\right)\)
\(\Rightarrow M=2^{100}-1\)