A Production Recurrence with a Wrong Guess
A small factory starts with 1 thousand units produced in week 1. Each week output is set by a_{n+1}=2a_n+n, where a_n is output in week n, in thousands of units. A planning assistant proposes the closed form a_n=2^n-n and uses it to forecast week 10. The floor manager notices that the assistant's formula gives a_2=2, but the recurrence gives a_2=3, so the forecast is suspect. The factory needs a correct closed form to decide whether weekly output will exceed 50 thousand units before week 12, and whether the weekly increase a_{n+1}-a_n will eventually be more than 1 thousand units.
1. Show the assistant's formula is wrong and derive the correct closed form. Check it at n=1 and n=2.
Read the worked answer
The proposed formula fails at n=2: it gives 2^2−2=2, while the recurrence gives a_2=2a_1+1=3.
To solve, try a_n=A·2^n+Bn+C. Substitution gives A·2^{n+1}+B(n+1)+C = 2(A·2^n+Bn+C)+n. Comparing coefficients of n and constants yields B=−1 and C=−1. Then a_1=2A−2=1 gives A=3/2.
Hence a_n=(3/2)2^n−n−1=3·2^{n−1}−n−1. Check: a_1=3−2=1, a_2=6−3=3.
2. Find the first week n with a_n>50 and verify your answer by computing the neighboring weeks.
Read the worked answer
Compute a_5=3·2^4−5−1=48−6=42, which is below 50. Compute a_6=3·2^5−6−1=96−7=89, which is above 50.
The recurrence gives a_{n+1}−a_n=a_n+n>0, so the sequence is increasing. Therefore the first week with output above 50 thousand units is n=6. This is before week 12.
3. Determine whether a_{n+1}-a_n>1 for all n≥1, and interpret the result for the factory's claim.
Read the worked answer
From the closed form,
a_{n+1}−a_n = [3·2^n−(n+1)−1] − [3·2^{n−1}−n−1] = 3·2^{n−1}−1.
For n=1 this is 2, and it grows with n. Hence a_{n+1}−a_n>1 for every n≥1, so the weekly increase is more than 1 thousand units from the first transition onward, not merely eventually. The factory's claim is true, and stronger than stated.