\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

Lời giải:

8A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^8-1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^{16}-1)(3^{16}+1)-4.3^{32}$

$=3^{32}-1-4.3^{32}$

$=-3.3^{32}-1=-3^{33}-1$
$\Rightarrow A=\frac{-3^{33}-1}{8}$

a) = \(\frac{127}{96}\)

b) = \(\frac{255}{256}\)

c) Mik bỏ nha

d) = \(\frac{1023}{512}\)

e) = \(\frac{2343}{625}\)

bạn có thể trả lời rõ ra được ko

a)

Dãy trên có số số hạng là:

( 20 - 1 ) : 1 + 1 = 20 ( số hạng )

Tổng của dãy trên là:

( 20 + 1 ) x 20 : 2 = 210 

Đáp số: 210

b)

Dãy trên có số số hạng là:

( 21 - 1 ) : 2 + 1 = 11 ( số hạng )

Tổng của dãy trên là:

( 21 + 1 ) x 11 : 2 = 121

Đáp số: 121

c) ( 2x - 1 ) x 2 = 13

2x - 1 = \(\dfrac{13}{2}\)

2x = \(\dfrac{15}{2}\)
\(x=\dfrac{15}{4}\)

32 x ( x - 10 ) = 32

( x - 10 ) = 1

x = 11

\(A=1+2+3+...+20\)

Số hạng:

\(\left(20-1\right):1+1=20\) (số hạng)

Tổng: \(\left(20+1\right)\cdot20:2=210\)

\(B=1+3+5+...+21\)

Số hạng:

\(\left(21-1\right):2+1=11\) (số hạng) 

Tổng: \(\left(21+1\right)\cdot11:2=121\)

\(\left(2x-1\right)\cdot2=13\)

\(\Rightarrow2x-1=\dfrac{13}{2}\)

\(\Rightarrow2x=\dfrac{15}{2}\)

\(\Rightarrow x=\dfrac{15}{4}\)

\(32\cdot\left(x-10\right)=32\)

\(\Rightarrow x-10=1\)

\(\Rightarrow x=11\)

`a,` Khoảng cách là : `1`

Số số hạng là : \(\dfrac{20-1}{1}+1=20\)

Tổng là : \(\dfrac{20+1\times20}{2}=\)`210`

`b,` Khoảng cách là : `2`

Số số hạng là : \(\dfrac{21-1}{2}+1=11\)

Tổng là : \(\dfrac{21+1\times11}{2}=121\)

\(c,\left(2x-1\right).2=13\\ \Rightarrow2x-1=\dfrac{13}{2}\\ \Rightarrow2x=\dfrac{13}{2}+1\\ \Rightarrow2x=\dfrac{15}{2}\\ \Rightarrow x=\dfrac{15}{2}:2\\ \Rightarrow x=\dfrac{15}{4}\)

\(32.\left(x-10\right)=32\\ \Rightarrow x-10=32:32\\ \Rightarrow x-10=1\\ \Rightarrow x=1+10\\ \Rightarrow x=11\)

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)

\(\frac{2+3+4+....+33}{2}\)

\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)

tớ không biết đâu

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

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

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

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

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

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

\(2A=3^{64}-1\Rightarrow A=\dfrac{3^{64}-1}{2}\)

Áp dụng đẳng thức : a^2 - 1 = (a + 1)(a - 1)
=> a + 1 = (a^2 - 1)/(a + 1)

Ta có: 3 + 1 = (3^2 - 1)/(3 - 1)
3^2 + 1 = (3^4 - 1)/(3^2 - 1)
3^4 + 1 = (3^8 - 1)/(3^4 - 1)
3^8 + 1 = (3^16 - 1)/(3^8 - 1)
3^16 + 1 = (3^32 - 1)/(3^16 - 1)
3^32 + 1 = (3^64 - 1)/(3^32 - 1)

(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
=(3^2 - 1)/(3 - 1).(3^4 - 1)/(3^2 - 1).(3^8 - 1)/(3^4 - 1).(3^32 - 1)/(3^16 - 1).(3^64 - 1)/(3^32 - 1)
=(3^64 - 1)/(3 - 1)
=(3^64 - 1)/2