Câu hỏi của Nguyễn Thiên Vy

Question

BÀI 1: a) 32<2^n<128          b) 2.16> hặc = 3^n < hoặc = 24^3          c) 9.27 < hoặc = 3^n < hoặc = 24^3BÀI 2:A=(1/2+1/2)^2+(1/2)^3+(1/2)^4+....+(1/2)^98+(1/2)^99

VRCT_Ran Love Shinichi · Accepted Answer

Bài 1::a) 32<2n<128=>25<2n<27=>n=6Bài 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 = 1Câu trả lời: Bài 1::a) 32<2n<128=>25<2n<27=>n=6Bài 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

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