# Sequence problem

K it's been a while and I cant remember how to figure this problem out without going doing tons of work.

Estimating that the world population was 0.9 billion in 1798, use this equation to estimate the population (in billions) in the years 1799, 1800, 1801, 1802, and 1900.

Using the equation y(n+1) = 1.03y(n)

I got all the years except 1900... I don't remember how to find this out the fast way.. please refresh my memory

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K it's been a while and I cant remember how to figure this problem out without going doing tons of work.

Estimating that the world population was 0.9 billion in 1798, use this equation to estimate the population (in billions) in the years 1799, 1800, 1801, 1802, and 1900.

Using the equation y(n+1) = 1.03y(n)

I got all the years except 1900... I don't remember how to find this out the fast way.. please refresh my memory
It bottles down to solving the difference equation y(n+1)-1.03y(n)=0
solving it we get $$y[n] = C1 * (1.03)^n$$ C1 is a constant
use initial condition at n =0 we have y(0) = 0.9 (in billions)
so C1 = 0.9
then the solution is $$y[n] = 0.9 * (1.03)^n$$ in billions
so year 1900 is n= 1900-1798+1 = 103
Then u get y = 18.9 Billion in the year 1900.

Dick
Homework Helper
It bottles down to solving the difference equation y(n+1)-1.03y(n)=0
solving it we get $$y[n] = C1 * (1.03)^n$$ C1 is a constant
use initial condition at n =0 we have y(0) = 0.9 (in billions)
so C1 = 0.9
then the solution is $$y[n] = 0.9 * (1.03)^n$$ in billions
so year 1900 is n= 1900-1798+1 = 103
Then u get y = 18.9 Billion in the year 1900.
Don't give the answer, ok? Let the poster find if for him/her self. It takes the fun out of it for them.

Don't give the answer, ok? Let the poster find if for him/her self. It takes the fun out of it for them.
sorry u are right.... I have avoided this and if u noticed recently I been giving hints and explanations more than solving it to the end (unless required..)

not even correct.