a=511/256

b=647/20

c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời

a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)

= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256

= 2 - 1/256

= 511/256

Câu b bạn có viết sai đề không vậy?

b, 3/1x4 + 3/2x5 + 3/3x6 + 3/4x7 + 1/5x8

= 3/4 + 3/10 + 3/18 + 3/28 + 1/40

= 1133/840

c,2/1x3 + 2/3x5 + 2/5x7+..+ 2/2001x2003 + 2/2003x2005

= ( 1 - 1/3) + ( 1/3 - 1/5) + ( 1/5 - 1/7) +...+ ( 1/2001 - 1/2003) + (1/2003 - 1/2005)

= 1 - 1/3 +  1/3 - 1/5 + 1/5 - 1/7 +...+  1/2001 - 1/2003 + 1/2003 - 1/2005

= 1 - 1/2005

= 2004/2005

A = 3/1.3 + 3/3.5 + 3/5.7 + 3/7.9 + ... + 3/97.99

A = 3/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + .... + 2/97 - 2/99

A = 3/2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 )

A = 3/2 . ( 1 - 1/99 )

A = 3/2 . 98/99

A = 49/33

b) dãy số không có quy luật==> bạn xem lại đề

c) \(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{50\times51}-\frac{1}{51\times52}\right)\)

\(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{51\times52}\right)=\frac{1}{2}\times\frac{2650}{5408}=\frac{1325}{5408}\)

\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)

* Bạn làm theo công thức và vẫn dụng câu b nhé.

A=1x2+2x3+3x4+...+49x50

3A= 3(1.2+2.3+3.4+...+49.50)

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3

3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)

3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51

3A= 49.50.51

A= 49.50.51/3=41650

B=1x3+3x5+5x7+...+99x101

B=1/1.3 +1/3.5 +...+1/99.101

2B=2/1.3 + 2/3.5 +...+2/99.101

2B=1-1/3+1/3-1/5+...+1/99-1/101

2B=1-1/101

2B=100/101

B=100/101:2=100/202

C=1x2x3+2x3x4+3x4x5+...+50x51x52
Nhân C với 4 ta được:
C x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+50x51x52x4
C x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 50x51x52x(53-49)
C x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... +49x 50x51x52 - 50x51x52x53
Sau khi cộng - trừ giản ước ta có : C x 4 = 50x51x52x53
C = 50x51x52x53 : 4 = 1756950

Lời giải:

$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}$

$=1-\frac{1}{99}=\frac{98}{99}$

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{99\times101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)

= 100/101