Question

[Nguyễn Thiên Vy]

BÀI 1: a) 32<2^n<128

          b) 2.16> hặc = 3^n < hoặc = 24^3

          c) 9.27 < hoặc = 3^n < hoặc = 24^3

BÀI 2:

A=(1/2+1/2)^2+(1/2)^3+(1/2)^4+....+(1/2)^98+(1/2)^99

Answer 1

[VRCT_Ran Love Shinichi]

Bài 1::

a) 32<2ⁿ<128

=>2⁵<2ⁿ<2⁷

=>n=6

Bài 2:Ta có :

          A = 1/2+(1/2)²+(1/2)³+...+ (1/2)⁹⁸+(1/2)⁹⁹+(1/2)⁹⁹

 => 1/2A = (1/2)²+(1/2)³+...+ (1/2)⁹⁸+(1/2)⁹⁹+(1/2)¹⁰⁰+(1/2)¹⁰⁰

 1/2B- A = [(1/2)²+(1/2)³+...+ (1/2)⁹⁸+(1/2)⁹⁹+(1/2)¹⁰⁰+(1/2)100] - [ 1/2+(1/2)²+(1/2)³+...+ (1/2)⁹⁸+(1/2)⁹⁹+(1/2)⁹⁹]

     -1/2A = [(1-2)²-(1/2)²]+[(1/2)³-(1/2)³]+...+[(1/2)⁹⁸-(1/2)⁹⁸]+[(1/2)⁹⁹-(1/2)⁹⁹]+[(1/2)¹⁰⁰+(1/2)¹⁰⁰-(1/2)⁹⁹] -1/2

     -1/2A = 0+0+...+0+0+0-1/2

     -1/2A = -1/2

=>       A = 1