ta co: 1/4=1/2(1+1) ;

2/6 = 1+1/2(2+1)..... 
=>ve trai =n/2(n+1).(n+1)/2(n+2).(n+2)/2(n+3)..... 
= (1/2)^36 
=> n= - 36

`lim [4.2^[n+1]-10^[n+2]]/[3.5^n-10^n]`

`=lim [2.(1/5)^[n+2]-1]/[3/25(1/2)^[n+1]-1/100]`

`=[-1]/[-1/100]=100`

m

a) 3ⁿ⁺¹ = 3⁴

=> n + 1 = 4

=> n = 4 - 1

=> n = 3

Vậy n = 3

b) 4.2ⁿ = 64

=> 2ⁿ = 64 : 4

=> 2ⁿ = 16 = 2⁴

=> n = 4

Vậy n = 4

\(3^{x+1}=4\)

\(\Rightarrow x+1=4\)

\(\Rightarrow x=4-1\)

\(\Rightarrow x=3\)

\(H=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}...\frac{30}{62}.\frac{31}{64}\)

\(\Rightarrow H=\frac{1.2.3...30.31}{2.2.2.3.2.4...2.31.2.32}=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}\)

(18+38+16*76-1)=(36*19+64*20-65)-x

=>(18+38+1216-1)=(684+1280-65)-x

=>1271=1899-x

=>1899-x=1271

x = 1899-1271

x=628

vậy x= 628

 ( 18 + 38 + 1216 - 1) = (  684 +1280 - 65) - x

 ( 56 + 1216 - 1) = ( 1964 - 65) - x

 1271 = 1899 - x

x = 1899 - 1271

x = 628 

\(\Leftrightarrow\dfrac{2\cdot3\cdot...\cdot31}{\left(2\cdot3\cdot...\cdot31\right)\cdot\left(2\cdot2\cdot...\cdot2\right)\cdot2^6}=2^x\)

=>2^x=1/2^36

=>x=-36