Question

1/4+1/8+1/16+1/32+1/64+1/128.

Answer 1

\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{32+16+8+4+2+1}{128}=\frac{63}{128}\)

Answer 2

Đặt A=1/4+1/8+1/16+1/32+1/64+1/128.

\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)

\(2A=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)

\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

\(A=\frac{1}{2}-\frac{1}{2^7}\)

Answer 3

A=1/4+1/8+1/16+1/32+1/64+1/128

Ax2-A=(1/4+1/8+1/16+1/32+1/64+1/128)x2-(1/4+1/8+1/16+1/32+1/64+1/128)

Ax2-A=(1/2+1/4+1/8+1/16+1/32+1/64)-(1/4+1/8+1/16+1/32+1/64+1/128)

Ax2-A=1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128

A=1/2-1/128

A=    63/128.

Answer 4

1/4+1/8+1/16+1/32+1/64+1/128 = 63,128

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Answer 5

63/128