P =2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10  

    = ( 2+2^2 ) + ( 2^3 + 2^4 ) + ( 2^5 + 2^6 ) + ( 2^7 + 2^8 ) + ( 2^9 + 2^10 )

   = 2.( 1+2 ) + 2^3.( 1+2 ) + 2^5.( 1+2 ) + 2^7.( 1+2 ) + 2^9.( 1 + 2 )

   = 2.3+2^3.3+2^5.3+2^7.3+2^9.3

   = ( 2+2^3+2^5+2^7+2^9).3 chia hết cho 3

=> P chia hết cho 3

câu 1 :19

câu 2:1

câu 3:3

câu 4:4

câu 5:có chia hết cho 3 vì tổng =2046

câu 1:19

câu 2:1

câu 3:3

câu 4:4

câu 5: có chia hết cho ba vì tổng = 2046

A = \(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+....+\frac{1}{2^{99}}\)

2²A = \(2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\)

=> 3A = \(2-\frac{1}{2^{99}}\)

=> A = \(\frac{2-\frac{1}{2^{99}}}{3}\)

a. \(A=2+2^2+2^3+2^4+2^5+...+2^{200}\)

\(2A=2^2+2^3+2^4+2^5+2^6+...+2^{201}\)

\(2A-A=2^{201}-2\)

b. \(Tacó:2^1=2;2^2=4;2^3=8;2^4=..6;2^5=2;2^6=4;...\)

Cứ dựa theo quy luật trên, ta có:

\(2^{4k+1}=..2;2^{4k+2}=...4;2^{4k+3}=...8;2^{4k}=..6\)

\(=>2^{201}=2^{4.50+1}\)

\(=>2^{101}=...2\)

\(=>2^{201}-2=..2-2\)\(=0\)

Vậy A có tân cùng là 0

Ta có: P=1+2+2²+2³+.........+2⁹⁹  (1)

=> 2P= 2+2²+2³+2⁴+........+2¹⁰⁰  (2)

Lấy (2)-(1) => 2P-P=(2+2²+2³+2⁴+.......+2¹⁰⁰)-(1+2+2²+2³+..........+2⁹⁹)

                 => P=2¹⁰⁰-1

a)10⁸.2⁸=(10.2)⁸=20⁸

b)10⁸:2⁸=(10:2)⁸=5⁸

c)25⁴.2⁸=(5²)⁴.2⁸=5^2.4.2⁸=5⁸.2⁸=(5.2)⁸=10⁸

a. 10 mũ 8 . 2 mũ 8 = 10 . 2 mũ 8 = 20 mũ 8

b. 10 mũ 8 : 2 mũ 8 = 10 : 2 mũ 8 = 5 mũ 8

c . 25 mũ 4 . 2 mũ 8 = 390625 . 64 = 25000000

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