### The Task

The sum of the squares of the first ten natural numbers is,

`1^2 + 2^2 + 3^3 + ... + 10^2 = 385`

The square of the sum of the first ten natural numbers is,

`(1 + 2 + 3 + ... + 10)^2 = 55^2 = 3025`

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is:

`3025 - 385 = 2840`

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

There are three solutions to this task:

- A straightforward one -> 14 lines
- A functional, "lispy" one -> 3 lines
- A mathematical one -> 2 lines.

### The straightfoward solution

This one is very simple, because we just do literally what is in the task description. We have three functions: `sum_squared`

, which sums up and squares the result, `squares_sum`

, which forms the square and then sums up, and `task_6`

which combines it to deliver the solution.

```
(de sum_squared (N)
(let Sum 0
(for I N
(inc 'Sum I) )
(* Sum Sum)) )
(de squares_sum (N)
(let Sum 0
(for I N
(inc 'Sum (* I I)) )
Sum ) )
(de task_6 (N)
(-
(sum_squared N)
(squares_sum N)) )
```

### The functional solution

The code above smells a little bit like repetition and redundancy. Redundant tasks in `sum_squared`

and `squares_sum`

:

- In both functions, we use a helper variable
`I`

to sum up from 0 to N. - in both functions, we increment a variable
`Sum`

.

Let's first try to improve the first point. Instead of `for I N`

, we could use a list, because we all know that PicoLisp has powerful list functions. With `range`

, we can create a list from 1 to `N`

.

```
: (setq L (range 1 10))
-> (1 2 3 4 5 6 7 8 9 10)
```

Next, we can use `apply`

to apply the `+`

function to *all elements on the list*.

```
: (apply + L)
-> 55
```

This is equivalent to the first 4 lines of our `sum_squared`

function above.

In the same way, we can use the `sum`

function to apply the `*`

function to *each element of a list* and return the numerical sum of the list elements. The syntax is `(sum 'fun 'lst ...)`

:

```
: (sum * L L)
-> 385
```

This single line is equivalent to our `squares_sum`

function.

In other words, the above code can be shortened down to:

```
(de task_6 (N)
(let (L (range 1 N) S (apply + L))
(- (* S S) (sum * L L)) ) )
```

Note: Depending on the usecase, you need to cosnider that the `apply`

function requires a lot of stack space. If you have the latest PicoLisp version installed, you will get a "stack overflow" error message if the numbers get too large.

```
: (task_6 100000)
-> 25000166664166650000
: (task_6 1000000)
!? (apply + L)
Stack overflow
```

### The mathematical approach

Now, the "lispy" solution already has many improvements, such as reducing redundancy and repetition. However, internally we still have a lot of looping and iteration going on, which causes some (potential) limitations on the applicable range - depending on the use case of course. So let's look for mathematical approaches to improve our solution further.

There is a famous story about young Gauss who allegedly finished the teacher's task to sum up all numbers from 1 to 100 within minutes. He did that with the so-called "Gaussian Sum Formula":

Accordingly, the square of the sum can be expressed as `((n^2 + n)/2 )^2)`

.

In order to find the sum of squares, we can use the formula for square pyramidal numbers:

Now we bring these two formulas together, extract `n`

and re-group the factors a little bit to make it more pretty. With some easy calculations, you can form it to the following identity:

In PicoLisp:

```
(de task_6 (N)
(/ ( * N (- (* N N) 1) (+ (* 3 N) 2)) 12) )
```

This solution don't needs any loops or iterations, which means that the computation time will stay almost constant.

You can find the source code of the examples here: