help me in this problem please: u_{n+1}=(4u_n+3)/(u_n+6)

[ahmed.25]

a sequence (Un) n∈ℕ defined with Un+1=(4Un+3)/(Un+6)

question:
-confirm that Un+1=4-((21)/(Un+6))

i want to start from this
u_n+1=(u_n+3)/(u_n+6)
to get to this
u_n+1=(4)-((21)/(u_n​+6))

solved by Jomo

 

[Deleted member 4993]

a suite (Un) n∈ℕ defined with Un+1=4Un+3/Un+6 → is it..... \(\displaystyle \frac{3}{U_n + 6}\)?

question:
-confirm that Un+1=4-(21/Un+6) → same question here

.

 

[Steven G]

a suite (Un) n∈ℕ defined with Un+1=(4Un+3)/(Un+6)

question:
-confirm that Un+1=4-((21)/(Un+6))

i want to start from this
u_n+1=(u_n+3)/(u_n+6)
to get to this
u_n+1=(4)-((21)/(u_n​+6))

Maybe getting a common denominator might help???

 

[Deleted member 4993]

a sequence (Un) n∈ℕ defined with Un+1=(4Un+3)/(Un+6)

question:
-confirm that Un+1=4-((21)/(Un+6))

i want to start from this
u_n+1=(u_n+3)/(u_n+6)
to get to this
u_n+1=(4)-((21)/(u_n​+6))

Hint: simplify

\(\displaystyle 4 - \dfrac{21}{x + 6} \ = \ ?\)

 

[Steven G]

a sequence (Un) n∈ℕ defined with Un+1=(4Un+3)/(Un+6)

question:
-confirm that Un+1=4-((21)/(Un+6))

i want to start from this
u_n+1=(u_n+3)/(u_n+6)
to get to this
u_n+1=(4)-((21)/(u_n​+6))

Start with u_n+1=(4)-((21)/(u_n​+6)) to get u_n+1=(u_n+3)/(u_n+6) and read it (ie write it) from the bottom to the top.

Or start with (u_n+3)/(u_n+6) = 4 - (4 - (u_n+3)/(u_n+6)). This way you have your 4 - (safe guard it with your life!) and what is in the () you'll have to show its (21)/(u_n​+6) by getting a common denominator)

 

[ahmed.25]

Start with u_n+1=(4)-((21)/(u_n​+6)) to get u_n+1=(u_n+3)/(u_n+6) and read it (ie write it) from the bottom to the top.

Or start with (u_n+3)/(u_n+6) = 4 - (4 - (u_n+3)/(u_n+6)). This way you have your 4 - (safe guard it with your life!) and what is in the () you'll have to show its (21)/(u_n​+6) by getting a common denominator)

thanks that worked