A sequence is generated by the recurrence relation $u_{n+1} = mu_n + 6$ where $m$ is a constant - Scottish Highers Maths - Question 9 - 2017
Question 9
A sequence is generated by the recurrence relation $u_{n+1} = mu_n + 6$ where $m$ is a constant.
(a) Given $u_1 = 28$ and $u_2 = 13$, find the value of $m$.
(b)
(... show full transcript
Worked Solution & Example Answer:A sequence is generated by the recurrence relation $u_{n+1} = mu_n + 6$ where $m$ is a constant - Scottish Highers Maths - Question 9 - 2017
Step 1
Given $u_1 = 28$ and $u_2 = 13$, find the value of $m$.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
From the recurrence relation, we can substitute n=1:
u2=mu1+6
Substituting the known values gives:
13=m⋅28+6
Now, we isolate m:
13−6=28m7=28mm=287=41
Thus, the value of m is ( \frac{1}{4} ).
Step 2
Explain why this sequence approaches a limit as $n \to \infty$.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The sequence approaches a limit because the absolute value of the coefficient m=41 satisfies the condition:
−1<m<1
That means the sequence is stable enough for the terms to converge as n increases.
Step 3
Calculate this limit.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the limit, we set the limit L of the sequence when n→∞. Thus:
L=mL+6
Substituting m=41 gives:
L=41L+6
To isolate L, rearranging yields:
L−41L=643L=6L=6⋅34=8
Therefore, the limit as n→∞ is 8.
Join the Scottish Highers students using SimpleStudy...
