What's the formula for the 365 day penny challenge?

Question

_{Not exactly a duplicate since this is answering a specific instance popular in social media.}

You might have seen the viral posts about "save a penny a day for a year and make $667.95!" The mathematicians here already get the concept while some others may be going, "what"? Of course, what the challenge is referring to is adding a number of pennies to a jar for what day you're on. So:

Day 1 = + .01
Day 2 = + .02
Day 3 = + .03
Day 4 = + .04

So that in the end, you add it all up like so:

1 + 2 + 3 + 4 + 5 + 6 + ... = 66795

The real question is, what's a simple formula for getting a sum of consecutive integers, starting at whole number 1, without having to actually count it all out?!

Answer 1

Have had a lot of friends ask about this lately, as it is all over FaceBook. The formula is actually quite simple:

(N (N + 1) ) / 2 where N = Highest value

Or Simply $\frac {n(n+1)}{2}$

Thus

365 (365 + 1) ) / 2 = 66795

Divide that by 100 (because there's 100 pennies in a dollar) and viola! $667.95

Now, this is an OLD math (think about 6th century BC), wherein these results are referred to as triangle numbers. In part, because as you add them up, you can stack the results in the shape of a triangle!

1 = 1
     *
1 + 2 = 3 
     *
    * *
1 + 2 + 3 = 6
     *
    * *
   * * *
1 + 2 + 3 + 4 = 10
     *
    * *
   * * *
  * * * *

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NoChance also has a fun story and answer to this question!

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A little info on his lesson: _{-{for the super nerdy!}-}

"...Carl Friedrich Gauss is said to have found this relationship in his early youth, by multiplying n/2 pairs of numbers in the sum by the values of each pair n+1. However, regardless of the truth of this story, Gauss was not the first to discover this formula, and some find it likely that its origin goes back to the Pythagoreans 5th century BC..." - wikipedia

"...The mathematical study of figurate numbers is said to have originated with Pythagoras, possibly based on Babylonian or Egyptian precursors. Generating whichever class of figurate numbers the Pythagoreans studied using gnomons is also attributed to Pythagoras. Unfortunately, there is no trustworthy source for these claims, because all surviving writings about the Pythagoreans are from centuries later. It seems to be certain that the fourth triangular number of ten objects, called tetractys in Greek, was a central part of the Pythagorean religion, along with several other figures also called tetractys. Figurate numbers were a concern of Pythagorean geometry. ... - wikipedia

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See? Fun stuff, numbers!

Answer 2

As others pointed out, the answer is $\frac {n(n+1)}{2}$. Here is an intuitive proof:

You can group first and last number whose sum is $n + 1$. The second and the second last have sum $n + 1$. If you continue like that, you will notice that all such pairs have sum $n + 1$ - that is, because the first one gets increased with $1$, the second in the pair - with $-1$. How many pairs are there? $\frac{n}{2}$. So total sum of pairs is $\frac {n(n+1)}{2}$

Answer 3

Let $S_1=1+2+3+....+(n-1)+n$,

and $S_2=n+(n-1)+(n-2)+...+2+1$, it should be clear that $S_1=S_2$

Add the two expression gives $S_1+S_2=(n+1)+(n+1)+...+(n+1)$ there are n terms, i.e. $2S_1=n(n+1)$ or $S_1=\frac{n(n+1)}{2}$

Answer 4

$1+2+3+...+(n-1)+n=\frac{1+2+3+...+(n-1)+n+1+2+3+...+(n-1)+n}{2}=\frac{(1+n)+(2+n-1)+(3+n-2)+...+(n+1)}{2}=\frac{(n+1)+(n+1)+(n+1)+...+(n+1) \left[ n \mathrm{-times}\right]}{2}=\frac{n(n+1)}{2}$

Answer 5

The real question is, what's a simple formula for getting a sum of consecutive integers, starting at whole number 1, without having to actually count it all out

While others have answered the question, I could not resist to reflect some history associated with the question.

The question you asked relates back to a famous mathematician Gauss the story sometimes referred to as "Gauss Punishment", goes like:

In elementary school in the late 1700’s, Gauss was asked to find the sum of the numbers from 1 to 100. The question was assigned as “busy work” by the teacher, but Gauss found the answer rather quickly by discovering a pattern. His observation was as follows:

1 + 2 + 3 + 4 + … + 98 + 99 + 100

Gauss noticed that if he was to split the numbers into two groups (1 to 50 and 51 to 100), he could add them together vertically to get a sum of 101.

1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50

100 + 99 + 98 + 97 + 96 + … + 53 + 52 + 51

1 + 100 = 101 2 + 99 = 101 3 + 98 = 101 . . . 48 + 53 = 101 49 + 52 = 101 50 + 51 = 101

Gauss realized then that his final total would be 50(101) = 5050.

The source of the above is mostly from The sum of the first 100 whole numbers

Another version goes like, he wrote the numbers as follows:

001 + 002 + 003 +...+ 098 + 099 + 100 = S

100 + 099 + 098 +...+ 003 + 002 + 001 = S

(100+1)+(100+1)+(100+1)+...+ (100+1)+(100+1)+(100+1)= 2S

The value $(100+1)$ is repeated $100$ times.

so we get:

$$100 * (100+1) = 2S$$

but we only want the value of $S$

$$s=\frac{100*(100+1)}{2}$$

Needless to say, the number 100 can be any positive integer and the method would work the same. It is amazing what goes in the mind of a kid who is very young!

Answer 6

As the growth is linear, the average amount is also the average of the first-day and last-day amounts, hence

$$365\times\frac{0.01+3.65}2.$$

Answer 7

Generically speaking, you can add up any evenly spaced set of numbers with the formula:

$(1+\frac{n_{2}-n_{1}}{i})(\frac {n_{2}+n_{1}}{2})$

Given that $i$ is the interval between each number, $n_1$ is the lower number, and $n_2$ is the higher number.

For example, if you started at \$5, then counted up two dollars at a time for a week (to \$17), you would end up with \$77.

$(1+\frac{17-5}{2})(\frac {17+5}{2})$

$(1+\frac{12}{2})(\frac {22}{2})$

$(1+6)(11)$

$7(11)$

$77$

The penny challenge follows the same formula:

$(1+\frac{365-1}{1})(\frac {365+1}{2})$

$(1+364)(183)$

$(365)(183)$

$66795$ (in pennies)

While the other answers focus on a specialized subset of the this formula, I thought I would provide a more general form of the formula that is useful in a variety of situations.

Answer 8

I'm surprised that no one has posted the solution by counting handshakes. If there are $N$ people in a room and everyone shakes hands with everyone else, how many handshakes are there?

Count this two ways. The room starts out with one person. Number $2$ arrives and shakes one hand. Number $3$ arrives and shakes two, and so on, so the total number is $1 + 2 + \cdots + (N-1)$.

Now count the handshakes another way. Each of the $N$ people shakes $N-1$ hands, but that counts each handshake twice, so the total is $N(N-1)/2$.

And do check out the second proof without words here: http://artofproblemsolving.com/wiki/index.php?title=Proofs_without_words#Summations

Answer 9

The arithmetic progression is the sequence of numbers such that the difference $d$ between the consecutive terms is constant. If the first term is $a_1$, the number of terms is $n$ and the last term is $a_n$, the whole sum

$$ S = \frac{n \cdot (a_1 + a_n)}{2} $$

where

$$ a_n = a_1 + (n - 1) \cdot d $$

In your example $a_1 = 0.01$, $d = 0.01$, $n = 365$, so

$$ a_{365} = 0.01 + (365 - 1) \cdot 0.01 = 3.65 $$

and

$$ S = \frac{365 \cdot (0.01 + 3.65)}{2} = 667.95 $$

If you want to use only $a_1$, $d$ and $n$ then the only one formula from both one above

$$ S = \frac{n \cdot (2 \cdot a_1 + (n-1) \cdot d)}{2} $$

$$ S = \frac{365 \cdot (2 \cdot 0.01 + (356-1) \cdot 0.01)}{2} = 667.95 $$