Question

A student decides she wants to save money to buy a used car, which costs $2600. She comes up with what she thinks is a very modest savings plan. She decides to save 2 cents the first day and double the amount she saves each day thereafter. On the second day she plans to save 4 cents, on the third day, 8 cents, and so on. Determine how long it will take her to save enough money to buy the car ?

Answer 1

Answer:

17 days

Step-by-step explanation:

as you see, 4=2^2

and 8=2^3

so we have a number sequence:

2+2^2+2^3+...+2^n=260000 (1)

multiply (1) by 2 we have:

2^2+2^4 +...+2^(n+1)=520000 (2)

(2) minus (1) we have:

2^(n+1)-2=260000

2^n*2=260002

2^n=130001

and we have 2^17=131072 >130001

so it will take her at least 17 days to buy the car

Answer 2

It will take her about 17 days to buy the car worth 260000 cents

If a student decides she wants to save money to buy a used car that cost $2600 (260,000 cents)

If she saves 2 cents the first day and doubles the amount thereafter, the sequence of savings will be:

2, 4, 8...

This sequence is geometric in nature

In order to determine how long it will take her to save 260,000, we will use the sum of a GP formula expressed as:

$S_n = \frac{a(r^n-1)}{r-1}$

Given the folowing

a = 2

r = 4/2 = 8/4 = 2

Sn = 260,000

Substitute into the formula the given parameters

$260000= \frac{2(2^n-1)}{2-1}\\260000/2=2^n-1\\130000 = 2^n - 1\\2^n = 130000 + 1\\2^n = 130001\\nlog 2=log130001\\n = \frac{log130001}{log2} \\n \approx 17$

This shows that it will take her about 17 days to buy the car worth 260000 cents

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