a: Sửa đề: \(P=2-2^2+2^3-2^4+\cdots+2^{99}-2^{100}\)

=>\(2P=2^2-2^3+2^4-2^5+\cdots+2^{100}-2^{101}\)

=>\(2P+P=2^2-2^3+2^4-2^5+\cdots+2^{100}-2^{101}+2-2^2+2^3-2^4+\cdots+2^{99}-2^{100}\)

=>\(3P=-2^{101}+2\)

=>\(P=\frac{-2^{101}+2}{3}\)

b: \(P=2-2^2+2^3-2^4+\cdots+2^{99}-2^{100}\)

\(=\left(2-2^2+2^3-2^4\right)+\left(2^5-2^6+2^7-2^8\right)+\cdots+\left(2^{97}-2^{98}+2^{99}-2^{100}\right)\)

\(=\left(2-2^2+2^3-2^4\right)+2^4\left(2-2^2+2^3-2^4\right)+\cdots+2^{96}\left(2-2^2+2^3-2^4\right)\)

\(=-10\left(1+2^4+\cdots+2^{96}\right)\) ⋮(-5)

sai đề r bạn ơi

A = 4 + 4² + 4³ + 4⁴ + ... + 4⁹⁹ + 4¹⁰⁰

A = ( 4 + 4² ) + ( 4³ + 4⁴ ) + ... + (4⁹⁹ + 4¹⁰⁰)

A = ( 4 + 4² ) + 4³(4 + 4² ) + .... + 4⁹⁹(4 + 4²)

A = 20 + 4³.20 + .... + 4⁹⁹.20

A = 20 ( 1 + 4³ + .... + 4⁹⁹ )

A = 4.5.(1 + 4^{3 + ...} + 4⁹⁹ ) ⋮ 5 ( đpcm )

\(A=4+4^2+4^3+4^4+...+4^{99}+4^{100}\)

\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{99}+4^{100}\right)\)

\(A=4\left(4+1\right)+4^3\left(4+1\right)+...+4^{99}\left(4+1\right)\)

\(=5\left(4+4^3+...+4^{99}\right)\Rightarrow A⋮5\)

Ta có : S = 1 - 3 + 3² - 3³ + 3⁴ - 3⁵ +...+ 3⁹⁸ - 3⁹⁹ 

      => 3S = 3 - 3² + 3³ - 3⁴ + 3⁵ - 3⁶ +...+ 3⁹⁹ - 3¹⁰⁰ 

Lấy 3S + S = (3 - 3² + 3³ - 3⁴ + 3⁵ - 3⁶ +...+ 3⁹⁹ - 3¹⁰⁰ ) + ( 1 - 3 + 3² - 3³ + 3⁴ - 3⁵ +...+ 3⁹⁸ - 3⁹⁹ )

          4S    = 3¹⁰⁰ + 1

=> \(S=\frac{3^{100}+1}{4}\Leftrightarrow3^{100}+1⋮4\) (vì sở dĩ tổng S là số nguyên) 

=> 3¹⁰⁰ : 4 dư 1 

Ta có:

A = 4 + 4²  + 4³ + 4⁴ + ... + 4⁹⁹ + 4¹⁰⁰

A = (4 + 4²) + (4³ + 4⁴) + ... + (4⁹⁹ + 4¹⁰⁰)

A = 4(1 + 4) + 4³(1 + 4) + ... + 4⁹⁹(1 + 4)

A = 4.5 + 4³.5 + ... + 4⁹⁹.5

A = 5.(4 + 4³ + ... + 4⁹⁹)

Vậy A chia hết cho 5

\(A=4+4^2+4^3+...4^{99}+4^{100}\)

\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{99}+4^{100}\right)\)

\(A=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{99}.\left(1+4\right)\)

\(A=4.5+4^3.5+..4^{99}.5\)

\(A=5.\left(4+4^3+...4^{99}\right)\)

\(\Rightarrow A⋮5\)

a: 10100

b: có, đáp số là: 0,8037156704

A có 100 số hạng

Tổng A :

( 100 + 1 ) x 100 : 2 = 5050 \(⋮\)5

=> A \(⋮\)5