How much money will i save with this?

Question

I am thinking of doing this:
I want to open an account.
On the first day i deposit a shilling.
on the second day i deposit 2 shillings
on the third day i deposit 3 shillings.
i continue with this trend for one year( 365 days)
each new day i will be depositing one shiiling more than what i deposited yesterday untill the last day of the year when i will deposit the last instalment of 365 shillings.
At the end of the year(365 days) how much money will i have in my account?

Answer 1

At the end of the year, you will have 66795 shillings or
3339 pounds 15 shillings.

Here's how to work it out :

Use the formula for the sum of an arithmetic progression (or series) to evaluate:

1 + 2 + 3 + ... + 363 + 364 + 365.

The sum = (1/2) n [2 a + (n - 1) d] where n is the number of terms and [2 a + (n - 1) d] is in fact the sum of the first and last terms. (With this understanding, there is no need to break this latter expression down further.] A short cut to this is the quotable sum for all integers from 1 to n; that is n*(n + 1) / 2.

Therefore you have (1/2)*365*366 shillings = 66795 shillings or 3339 pounds 15 shillings --- quite a healthy savings for the year!

Live long and prosper --- which you're well on the way to doing.

Answer 2

1 + 2 + 3 ... + 363 + 364 + 365
+365+364+363 ....+ 3 + 2 + 1
= 366 + 366 ... 366

That's 366*365. Since we added an equation to itself twice, we can divide it by two to get the sum. So it's 183*365, or 66795

Answer 3

You are adding 1 shilling everyday.
So the constant amount that you make will be only for 183 days(app) i.e 365/2.
So if you do this calculation your saving at the end of 365th day will be 66795
Cal: 365(1/2) * 365 days

Answer 4

You are working with something called triangular numbers

The sum of all consecutive natural numbers from 1 to n (1+2+3+...+(n-1)+n) follows the general formula:

n(n+1)/2 or (n^2+n)/2 ----> remember this formula!

365(366)/2 = 66 795

Answer 5

a buck tooth and 2 indian nickels

Answer 6

Is the account interest-bearing?

Answer 7

it would be 66795