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# Money math problem

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If I go to a bank and deposit 1¢ into an interest bearing account and every month the amount in the interest bearing account doubles, how long in years and months will it take for my interest bearing account to get to at or just over \$1.000.000?  Please show step by step instructions on how to get to the answer.

Apr 10, 2019

#1
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1, 2, 4, 8, 16.......etc.

Use the Geometric Series Formula to sum them up:

Sum =F x [1 - R]^N / [1 - R], where F = First term, R=Common ratio, N =Number of terms.

\$1,000,000 x 100 = 1 x 2^N - 1

100,000,000 cents = 2^N - 1

100,000,001 = 2^N               Take the log of both sides

8 = N * log(2)

N = 8 / log(2)

N = 26.575 months = 27 month(rounded)

2^27 - 1 =134,217,727 cents / 100 =\$1,342,177.27 - dollars.

Note: The above is the same as:

FV =PV * [1 + 100%]^N - 1

\$1,000,000 =1 cent x 2^N -1

100,000,000 cents =2^N - 1 - and you  would solve for N as above.

Apr 10, 2019
#2
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Nice Job you are pretty smart you shoukd register it is completly free...

Nickolas  Apr 10, 2019
#3
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https://web2.0calc.com/questions/welcome-to-web2-0calc_14

Nickolas  Apr 10, 2019
#4
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Pretty straighforward:

2^n = 1 000 000 00

n log2 = 8

n = 26.57 months

Apr 10, 2019