A=1+3^2+3^4+...+3^100
Tính A? Trả lời giúp với ạ
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Do a chia 3 dư 2 nên a = 3k + 2 (k ∈ ℕ)
⇒ a² - 1 = (3k + 2)² - 1
= (3k)² + 2.3k.2 + 2² - 1
= 9k² + 12k + 3
= 3(3k² + 4k + 1) ⋮ 3
Vậy (a² - 1) ⋮ 3
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
Bài 4 :
A = 2 + 22 + 23 + 24 + ... + 2100
2A = 22 + 23 + 24 + 25 + ..... + 2101
2A - A = ( 22 + 23 + 24 + 25 + ..... + 2101 ) - ( 2 + 22 + 23 + 24 + ... + 2100 )
A = 2101 - 2
* Bài 5 bạn đợi chút ạ !!!
(Cách làm thì để mình nhắn riêng nhé)
Bài 4 :
A = 2 + 22 + 23 + ... + 2100
2A = 2.(2 + 22 + 23 +.....+ 2100)
2A = 22 + 23 + 24 + ... + 2101
A = 2101 - 2
Bải 5 :
A = 1 + 2 + 22 +.. + 24
2A = 2(1+2+22+ 23 + 24)
2A = 2 + 23 + 24 + 25
A = 25 - 1
=> A = B
b) C = 3 + 32 + .. +3100
3C = 3(3 + 32 + .. + 3100)
3C = 32 + ... + 3100
2C = 3101 - 3
C = (3101-3) : 2
=> C = D
\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)=\frac{1}{2}-\frac{13}{12}=\frac{-7}{12}\)
\(\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)=\frac{1}{24}-\frac{-5}{24}=\frac{1}{4}\)
mk tính ra rồi nhé còn lại bạn tự so sánh đi ak
a) (1+3x)^4 = 256
=> (1+3x)^4= 4^4
=> 1+3x=4
=> 3x=3
=> x=1
b) \(\frac{3^{15}.29+3^{15}.88}{3^{13}.81}=\frac{3^{15}.\left(29+88\right)}{3^{13}.81}=\frac{3^{15}.117}{3^{13}.81}=\frac{3^2.13}{9}=13\)
\(x=\dfrac{5}{a-1}< 0\)
\(\Leftrightarrow a-1< 0\Leftrightarrow a< 1\left(1\right)\)
Và \(x=\dfrac{5}{a-1}\in Z\)
\(\Rightarrow a-1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\left(2\right)\)
\(\Rightarrow a\left\{2;0;6;-4\right\}\)
\(\left(1\right),\left(2\right)\Rightarrow a\in\left\{-4;0\right\}\)
Sửa đề:
\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{50}\)
1-2+3-4...+99-100
= (1-2)+(3-4)...+(99-100)
= -1.(100/2)
= -50
A=1+3^2+3^4+...+3^100
-> 3^2A=9A=3^2+3^4+3^6+....+3^102
-> 9A-A=3^102-1( chỗ này mik làm tắt vì mỏi tay)
-> 8A=3^102-1
->A=\(\frac{3^{102}-1}{8}\)