## Starting with the number $1$ and moving to the right in a clockwise direction a $5$ by $5$ spiral is formed as follows:

21 22 23 24 25

20 7 8 9 10

19 6 1 2 11

18 5 4 3 1217 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is $101$.

What is the sum of the numbers on the diagonals in a $1001$ by $1001$ spiral formed in the same way?

### First, let’s notice a pattern from the smaller spiral. For the 5×5 grid, the top right corner, 25, is 5^2. Getting to the next highest number in the diagonal (going counter-clockwise), we subtract 4 (21, then 17, then 13). It seems to be 4 less than the previous one because each side of the square is 4 units longer than the previous square.

Then it’s important to observe that there are 4 corners in every layer apart from the center. For a 5×5 grid there are 2 layers (1, and then 3 to 5), for a 7×7 grid there are 3 layers (1, 3 to 5, and 5 to 7), and so on.

For an n x n grid, the sum of the four corners of the last layer is 4n^2 – 6n + 6.

Here 1 is at 0th layer.

Let’s add up the corners for the first several layers:

– For the layer 1 (3×3), n = 3. The sum = 4(3^2) – 6(3) + 6 = 24

– For the layer 2 (5×5), n = 5. The sum = 4(5^2) – 6(5) + 6 = 76

– For the layer 3 (7×7), n = 7. The sum = 4(7^2) – 6(7) + 6 = 160

And so on, and at last,

– For the layer 500 (1001×1001), n = 1001. The sum = 4(1001^2) – 6(1001) + 6.

So the sum of the numbers on the diagonals in a 1001 by 1001 spiral is 1 + 24 + 76 + 160 + … + [4(1001^2) – 6(1001) + 6].

This is essentially the sum of the series 1, 24, 76 … upto 500 terms.

To find it easier we can use the formula for the sum of an arithmetic series. An arithmetic series is a sequence of numbers in which each term after the first is obtained by adding a constant difference to the preceding term. For example, 1, 3, 5, 7, 9, 11, 13, 15, 17 is an arithmetic series with a common difference of 2.

The nth term of this series is 4n^2 – 6n + 6. The sum of first ‘n’ terms of an arithmetic series is given by the formula:

Sum = n/2 * [first_term + nth_term]

Here, the first term = 1 (n = 0) and last term = 4(1001^2) – 6(1001) + 6.

and n = 500 (there are 500 terms)

So, substituting these values into the formula we get:

Sum = 500/2 * [1 + 4(1001)^2 – 6(1001) + 6]

= 250 * [1 + 4*1002001 – 6006 + 6]

= 250 * [1003002]

= 250100500.

So, the sum of all numbers on the diagonals in a 1001×1001 spiral is 250100500.

##### More Answers:

$1000$-digit Fibonacci NumberReciprocal Cycles

Quadratic Primes