perplexus.info :: Just Math : Recurrent n's

Recurrent n's (Posted on 2024-06-02)

Let f₀(x)= x/(3x+2)
and, f_n(x)= f₀(f_n-1(x)), n≥1

Given that
f₂₀₁₈(x)= x/(Ax+B)
Determine the value of 3B-A

Note: Adapted from a problem appearing in Round 1 of Singapore Mathematical Olympiad Open 2018.

Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Solution: (Hide)

f_1(x) = f_0(f_0(x) = f_0(x/(3x+2))
= x/(3x+2)/(3*(x/3x+2)+2)
= x/9x+4
f_2(x) = f_0(f_1(x)) = f_0(x/9x+4)
= (x/9x+4)/(3*(x/9x+4)+2) = x/(21x+8)
Let f_n(x) = x/(a_n*x+b_n)
Then, a_0 = 3, a_1= 9, a_2=21
a_n = 3+2*a_(n-1)
=> a_(n+3) = 2(a_(n-1)+3)
Let u_n= a_n+3
=> u_n= 2u_(n-1)*u_n= a_0+3
=> u_n = 3*2*(2^n)= 3*2^(n+1)
=> a_n +3 = 3*2^(n+1)
=> a_n =3(2^(n+1)-1) and b _n = 2^(n+1)
3B -A = 3* 2^(n+1)- 3(2^(n+1)-1) = 3

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: No Subject	K Sengupta	2024-06-04 11:59:17
No Subject	dotota123	2024-06-04 09:34:45

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