Ask Me Help Desk - Arithmetic progression?

arithmetic progression?

if you put $1 in your piggy bank at the start of a year, another $2 the second day, another $3 the third day, and continue saving this way for a year, what will be the sum at the end of a year of 365 days

$no of days * average paid in per day + average paid in per day$

$365 \frac{365}{2} + \frac {365}{2} = $66795$

Quote:

Originally Posted by Capuchin

$\text{no of days * average paid in per day + average paid in per day}$

$365 \frac{365}{2} + \frac {365}{2} = $66795$

Hey Capuchin. Just in case you're interested. When typing text in LaTex, exclose it in [text]

Just in case you wasn't aware of that.

Capuchin, the way you wrote it doesn't seem straightforward to me. Do you have an explanation that makes sense? Clearly it is correct but I couldn't explain to someone why you'd multiply by the average then add the average. I was reminded of the explanation that I'd heard before and a fun story of Gauss here: Gauss and Series Summation

my description to the OP would be as follows:

instead of doing
1+2+... +364+365
if we note that the sum of the 1st and last, 2nd and next to last, and so on, all add to the same we will simplify our problem. If we add the nth number to the (365-n)+1st number we'll get the same sum each time (365+1=366). We'll be doing 365 sums. We'll have included each of the 365 numbers twice though so we'll have to divide by 2 at the end.

Code:

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

so we have 365 sums of (365+1)=366 but we have to divide by 2 since we used each number twice.
so our sum is 365*366/2 = 66795
the general formula for the sum of the numbers from 1 to N is:
$<br /> \sum_{n=1}^N n = N(N+1)/2<br />$

if you have any questions ask them!