1/128 * 2 = 1/64

1/64 * 2 = 1/32

1/32 * 2 = 1/16

...

1/4 * 2 = 1/2

1/2 * 2 = 1

Mà A chỉ có một số hạng 1/128 nên tính ra được 127/128

Bạn vào:câu hỏi của :Vũ Ngân Hà -olm

\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)

\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)

\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)

1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

a)

Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:

2/9+6/27+8/36+12/54+16/72+18/81=

2/9+2/9+2/9+2/9+2/9+2/9=

2/9*6=

12/9=

4/3

Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3

b)

Ta có:

1-2/5=3/5

1-2/7=5/7

1-2/9=7/9

...

1-2/99=97/99

Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=

3/5*5/7*7/9*...*97/99=

(3*5*7*...*97)/(5*7*9*...*99)=

3/99=

1/33

Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33

c)

Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:

S=1/2+1/4+1/8+1/16+...+1/1024

S*2=1+1/2+1/4+1/8+...+1/512

S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)

S=1-1/1024

S=1023/1024

Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024

Cảm ơn bạn nhé!

 dãy số trên là dãy số cách đều.
a) = sốcác số = ( 96-1 ) : 5 + 1= 20 số

    tổng câc số : (96 + 1 )x 20 : 2 = 970

b = dãy số từ 2 đến 100 là dãy số cách đều:

     số các số : (100 -2) : 2 + 1 =50 số

     tổng : (100 + 2) x 50 : 2 + (1+101) = 2652

a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)

\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)

\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)

b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)

\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)

\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)

=-3-1+1

=-3

c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)

\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)