\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)

\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-1\)

Ta đặt:    \(P=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

=> \(2P=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

=> \(2P-P=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

=> \(P=1-\frac{1}{64}\)

Mà    \(S=P-1\)

=> \(S=1-\frac{1}{64}-1=-\frac{1}{64}\)

Vậy \(S=-\frac{1}{64}\)

N = 3/2 + 5/4 + 9/8 + 17/16 + 33/32 + 65/64 - 7

N = (1 + 1/2) + (1 + 1/4) + (1 + 1/8) + (1 + 1/16) + (1 + 1/32) + (1 + 1/64) - 7

N = (1 + 1 + 1 + 1 + 1 + 1 ) + (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) - 7

N = 6 - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) - 7

Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64

2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

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

A = 1 - 1/64

N = 6 - (1 - 1/64) - 7

N = 6 - 1 + 1/64 - 7

N = 5 + 1/64 - 7

N = -2 + 1/64

N = -128/64 + 1/64

N = -127/64

bạn soyen_Tiểu bàng giải sao dòng 3 đang cộng đến dòng 9 lại chuyển thành trừ vậy

\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{31}{32}+\frac{65}{64}-7\)

\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+.....+\left(1+\frac{1}{64}\right)-7\)

\(=6+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-7\)

\(=.....................\)

\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\) \(7\)

\(=2\frac{3}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)

\(=3\frac{7}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)

\(=4\frac{15}{16}+\frac{33}{32}+\frac{65}{64}-7\)

\(=5\frac{31}{32}+\frac{65}{64}-7\)

\(=6\frac{63}{64}-7\)

\(=-\frac{1}{64}\)

Mình có cách như vậy thôi nhé, vì mình mới học lớp 5

\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)

\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{16}\right)+\left(1+\frac{1}{32}\right)+\left(1+\frac{1}{64}\right)-7\)

\(S=\left(1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-7\)

\(S=6+\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+....+\left(\frac{1}{32}-\frac{1}{64}\right)\right]-7\)

\(S=6+\left(1-\frac{1}{64}\right)-7\)

\(S=6+\frac{63}{64}-7\)

\(S=\frac{447}{64}-7=-\frac{1}{64}\)

s=1,5+1,25+1,125+1,0625+1,03125+1,015625-7=-0,015625

\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)

\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)

\(S=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-1\)

\(S+1=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)

\(2\left(S+1\right)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)

\(2\left(S+1\right)-\left(S+1\right)=S+1=1-\frac{1}{2^6}=\frac{63}{64}\)

\(S=\frac{63}{64}-1\)

A=(\(\dfrac{1}{3}-\dfrac{1}{3}\))\(+\left(\dfrac{3}{5}+\left(\dfrac{-3}{5}\right)\right)+\left(\dfrac{-5}{7}+\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)\)\(+\left(\dfrac{-11}{13}-\dfrac{9}{11}\right)\)

A\(=0+0+0+0+\dfrac{-238}{143}\)

A\(=\dfrac{-238}{143}\)

\(B=\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{4}\right)+\left(1+\dfrac{1}{8}\right)+\left(1+\dfrac{1}{32}\right)+\left(1+\dfrac{1}{64}\right)-7\)

\(B=\left(1+1+1+1+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)-7\)

\(B=6+\dfrac{63}{64}-7\)

\(B=-1+\dfrac{63}{64}\)

\(B=\dfrac{-1}{64}\)

3/2 + 5/4 + 9/8 + 17/6 + 33/32 + 65/2 - 7 

=(3/2+5/4+33/32+65/2) + 9/8+17/6 -7

= 1161/32 + 95/24 - 7

= 3191/26

3/2 + 5/4 + 9/8 + 17/6 + 33/32 + 65/2 - 7 = \(\frac{3191}{96}\)