Question

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A=2+2^2+2^3+2^4.....+2^100

B=1+3+3^2+3^2....+3^2009

C=1+5+5^2+5^3+...+5^1998

D=4+4^2+4^3+...+4^n

Answer 1

Ta có:

A=2+2^2+2^3+2^4+.....+2^100

=> 2A=2^2+2^3+...+2^101

=> 2A-A=A=(2^2+2^3+...+2^101)-(2+2^2+2^3+2^4.....+2^100)

=> A=2^2+2^3+...+2^101-2-2^2-...-2^100

=> A=2^101-2

B=1+3+3^2+3^2+....+3^2009

=> 3B=3+3^2+3^2+....+3^2010

=> 3B-B=2B=3+3^2+3^2....+3^2010-1-3-3^2-3^2-....-3^2009

=> 2B=3^2010-1

=> B=(3^2010-1)/2

C=1+5+5^2+5^3+...+5^1998

=> 5C=5+5^2+5^3+...+5^1999

=> 5C-C=4C=5+5^2+5^3+...+5^1999-1-5-5^2-5^3-...-5^1998

=> 4C=5^1999-1

=> C=(5^1999-1)/4

D=4+4^2+4^3+...+4^n

=> 4D=4^2+4^3+...+4^n+1

=> 4D-D=3D=4^2+4^3+...+4^n+1 - 4-4^2-4^3-...-4^n

=> 3D=4^n+1 - 4

=> 3D=\(\frac{4^{n+1}-4}{3}\)

Answer 2

Ta có : \(A=2+2^2+2^3+.....+2^{100}\)

\(2A=2+2^2+2^3+.....+2^{101}\)

\(2A-A=2^{101}-2\)

\(A=2^{101}-2\)

Answer 3

Ta có:

  A =2+2²+2³+2⁴+...+2¹⁰⁰

2A =2x(2+2²+2³+2⁴+...+2¹⁰⁰)

2A=    2²+2³+2⁴+2⁵+...+2¹⁰⁰+2¹⁰¹

 -A=2+2²+2³+2⁴+2⁵+...+2¹⁰⁰

  A=2¹⁰¹-2

     Vậy A=2¹⁰¹-2