Algebra

Progression & Series

Medium

If the population of a town is $p$ in the beginning of any year then it becomes $3+2 p$ in the beginning of the next year. If the population in the beginning of $2019$ is $1000$ then the population in the beginning of $2034$ will be

$(1003)2^{15} - 3$

$(997)2^{14}+ 3$

$(1003)^{15}+ 6$

$(997)^{15}-3$

✅ Correct Option: 1

We're given that if a town's population is p at the beginning of any year, it becomes 3 + 2p at the beginning of the next year.

Starting population in 2019: 1000

We need to find: Population in 2034

The population follows this pattern:

Population(next year) = 3 + 2 × Population(current year)

Let's trace through the first few years:

2019: Population = 1000

2020: Population = 3 + 2(1000) = 3 + 2000 = 2003

2021: Population = 3 + 2(2003) = 3 + 4006 = 4009

2022: Population = 3 + 2(4009) = 3 + 8018 = 8021

Let's examine these results more carefully:

2020: 2003 = 2006 - 3 = 2 × 1003 - 3

2021: 4009 = 4012 - 3 = 4 × 1003 - 3 = $2^2 × 1003 - 3$

2022: 8021 = 8024 - 3 = 8 × 1003 - 3 = $2^3 × 1003 - 3$

Pattern Discovery: After n years from 2019, the population is $2^n × 1003 - 3$

Why 1003? Notice that 1003 = 1000 + 3, where 1000 is our starting population.

Key Insight: This recurrence relation P(n+1) = 3 + 2P(n) can be solved using a substitution.

If we let Q(n) = P(n) + 3, then:

Q(n+1) = P(n+1) + 3

= (3 + 2P(n)) + 3

= 6 + 2P(n)

= 2(P(n) + 3)

= 2Q(n)

This means Q(n) grows by a factor of 2 each year!

Since P(2019) = 1000, we have Q(2019) = 1000 + 3 = 1003

Therefore: Q(2019 + k) = 1003 × $2^k$

Converting back: P(2019 + k) = Q(2019 + k) - 3 = $1003 × 2^k - 3$

From 2019 to 2034: k = 2034 - 2019 = 15 years

Population in 2034 = $1003 × 2^{15} - 3$

We don't need to calculate $2^{15}$ explicitly - the answer is $2^{15} × 1003 - 3$

When dealing with recurrence relations of the form P(n+1) = a + bP(n):

Try the substitution Q(n) = P(n) + c where c = $\frac{a}{1-b}$

This converts it to a geometric sequence: Q(n+1) = bQ(n)

Solve the geometric sequence and convert back

Final Answer: $2^{15}(1003) - 3$

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If the population of a town is ppp in the beginning of any year then it becomes 3+2p3+2 p3+2p in the beginning of the next year. If the population in the beginning of 201920192019 is 100010001000 then the population in the beginning of 203420342034 will be

Solution

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If the population of a town is $p$ in the beginning of any year then it becomes $3+2 p$ in the beginning of the next year. If the population in the beginning of $2019$ is $1000$ then the population in the beginning of $2034$ will be