a)\(A=1+3+3^2+...+3^{2018}\)

\(\Rightarrow3A=3.\left(1+3+3^2+...+3^{2018}\right)\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2019}\)

\(\Rightarrow3A-A=3+3^2+3^3+...+3^{2019}-\left(1+3+3^2+...+3^{2018}\right)\)

\(\Rightarrow2A=3^{2019}-1\)

\(\Rightarrow A=\frac{3^{2019}-1}{2}\)

b) \(B=5+5^2+...+5^{2017}\)

\(\Rightarrow5B=5^2+5^3+...+5^{2018}\)

\(\Rightarrow5B-B=5^2+5^3+...+5^{2018}-5-5^2-...-5^{2017}\)

\(\Rightarrow4B=5^{2018}-5\)

\(\Rightarrow B=\frac{5^{2018}-5}{4}\)

a,A=1+3+3²+...+3²⁰¹⁷

3A=3+3²+3³+...+3²⁰¹⁸

3A-A=3²⁰¹⁸-1

2A=3²⁰¹⁸-1

A=(3²⁰¹⁸-1):2

Giải:

a) Đặt:

\(A=1+2^2+2^3+2^4+...+2^{2018}\)

\(\Leftrightarrow2A=2+2^3+2^4+2^5+...+2^{2019}\)

\(\Leftrightarrow2A-A=\left(2+2^{2019}\right)-\left(1+2^2\right)\)

\(\Leftrightarrow A=2+2^{2019}-1-2^2\)

\(\Leftrightarrow A=2+2^{2019}-5\)

\(\Leftrightarrow A=2^{2019}-3\)

Vậy \(A=2^{2019}-3\).

b) Đặt:

\(B=1+5+5^2+5^3+...+5^{2017}\)

\(\Leftrightarrow5B=5+5^2+5^3+5^4+...+5^{2018}\)

\(\Leftrightarrow5B-B=5^{2018}-1\)

\(\Leftrightarrow4B=5^{2018}-1\)

\(\Leftrightarrow B=\dfrac{5^{2018}-1}{4}\)

Vậy \(B=\dfrac{5^{2018}-1}{4}\).

Chúc bạn học tốt!

a)A= 1 + 2²+2³ + 2⁴ +....+2²⁰¹⁸

2A = 2² + 2³ + 2⁴ +......+2²⁰¹⁸ + 2²⁰¹⁹

_

A= 1 + 2²+2³ + 2⁴ +....+2²⁰¹⁸

A= 2^{2019 _{- 1}}

a) \(A=2+2^2+2^3+...+2^{2017}\)

\(A=2\left(1+2^1+2^2+...+2^{2016}\right)\)

\(A=2.\dfrac{2^{2016+1}-1}{2-1}\)

\(A=2.\left(2^{2017}-1\right)=2^{2018}-2\)

Câu b bạn xem lại đề

a)

`A = 2 + 2^2+ ... + 2^2017`

`=> 2A = 2^2 + 2^3 + ... + 2^2018`

`=> 2A - A = (2^2 + 2^3 + ... + 2^2018) - (2 + 2^2 + ... +2^2017)`

`=> A         = 2^2018 - 2`

`B = 1 + 3^2 + ... + 3^2018`

`=> 3^2B = 3^2 + 3^4 + ... + 3^2020`

`=> 9B-B =(3^2 + 3^4 + ... + 3^2020) - (1 + 3^2 + ... + 3^2018`

`=> 8B     = 3^2020 - 1`

`=> B       = (3^2020 - 1)/8`

`C = 5 + 5^2 - 5^3 + ... + 5^2018`

`=> 5C = 5^2 + 5^3 - 5^4 + ... +5^2019`

`=> 5C + C = ( 5^2 + 5^3 - 5^4 + ... 5^2019) + (5 + 5^2 - 5^3 + ... + 5^2018)`

`=> 6C = 55 + 5^2019`

`=> C  = (5^2019 + 55)/6`

S=1+4+4 mũ 2+ 4 mũ 3 +....+ 4 mũ 2017

4S=4+ 4 mũ 2+ .....+4 mũ 2018

4S-S= (4+4 mũ 2+ 4 mũ 3+ ....+ 4 mũ 2018) - (1+4+4 mũ 2+ ......+ 4 mũ 2017)

S=4 mũ 2018 - 1

\(S=1+4+4^2+4^3+...+4^{2017}\)

\(4S=4+4^2+...+4^{2018}\)

\(4S-S=\left(4+4^2+...+4^{2018}\right)-\left(1+4+4^2+4^3+...+4^{2017}\right)\)

\(S=4^{2018-1}\)

\(S=4^{2017}\)

Ta có: \(A=\frac{2^{2017}+2}{2^{2017}+3}=1-\frac{1}{2^{2017}+3}\)

        \(B=\frac{2^{2017}+1}{2^{2017}+2}=1-\frac{1}{2^{2017}+2}\)

Vì \(\frac{1}{2^{2017}+3}< \frac{1}{2^{2017}+2}\) nên \(1-\frac{1}{2^{2017}+3}>1-\frac{1}{2^{2017}+2}\)

hay A > B

A>B bạn nhé

bạn viết lại đề đc ko bạn:>,ko hỉu đề

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Cộng à bn

\(A=2^0+2^1+2^2+...+2^{2010}\)

\(2A=2^1+2^2+2^3+...+2^{2021}\)

\(2A-A=\left(2^1+2^2+2^3+...+2^{2021}\right)-\left(2^0+2^1+2^2+...+2^{2020}\right)\)

\(A=2^{2021}-1\)

Ta có: \(A=100^2+200^2+300^2+...+1000^2\)

\(=100^2\cdot\left(1+2^2+3^2+...+10^2\right)\)

\(=100^2\cdot385=3850000\)

3800

3850