Question

Jerry plans on saving money for a trip by putting $1 in a savings account the first month and then adding

Answer 1

Answer:

1548

Step-by-step explanation:

To check your equation lets write out some terms of the sequence

 

1st month = 1,  total in account is 1

 

2nd month 1 + 1 = 2,  total in account is 2

 

3rd month 2 + 2 = 4,  total in account is 4

 

4th month 4 + 4 = 8, total in account is 8

 

5th month 8 + 8 = 16,  total in account is 16

 

So the sequence we have is

 

{1, 1+1, 2 + 2, 4 + 4, 8 + 8,...}→{1, 2, 4, 8, 16,...}

 

As I see it this is the sequence of partial sums so he does not have 1 + 2 =3 after two months.  He did not deposit two dollars for the second month.  He only puts enough money in the account to double what was already there.  So the summation equation is 

 

st = 2(t - 1),  t ≥ 1

 

Set this equation less than or equal to 1200 and solve for t

 

1200 ≤ 2(t - 1)

 

Take the log of each side

 

log(1200) ≤ log(2(t - 1))

 

Use the power rule

 

log(1200) ≤ (t - 1)log(2)

 

divide each side by log(2) and add 1

log(1200)/log(2) ≤ t - 1

 

log(1200)/log(2) + 1 ≤ t

 

t ≥ 11.23, rounding up gives t = 12

 

so he can contribute to the account for 12 months

 

At 12 months he has

 

s12 = 2^11 = 2048,  1024 in account and 1024 that he deposits

 

What left in the account is 2048 - 500 = 1548

Answer 2

9514 1404 393

Answer:

  s(t) = 2^(t-1)

Step-by-step explanation:

The amount in the account is the sum of the series ...

  1 + 1 +2 +4 +8 +16 +...

We note that terms after the first are a geometric sequence with a first term 1 and a common ratio 2. The sum of such a sequence is ...

  Sn = a1×(r^n -1)/(r -1) . . . . where n is the number of terms being summed

For our values, this simplifies to ...

  Sn = 1×(2^n -1)/(2 -1) = 2^n -1

We notice here that the account balance is 1 more than this sum, and that n = t-1, so the balance a(t) is ...

  s(t) = 1 +(2^(t -1) -1)

  s(t) = 2^(t -1) . . . . . . the function showing how much Jerry has