The Year 2025

This interesting set of facts about 2025 was passed on to me by a friend of mine, and contains some unexpected relationships between these integer sets.

Sum of the cubes

It starts off by noticing that 2025 can be expressed as the sum of cubes:

2025 = \sum_{i = 1}^{9} i^{3}

In general, the sum of the first n cubes can be written like this:

C_{n} = \sum_{i = 1}^{n} i^{3} = 1^{3} + 2^{3} + 3^{3} + ... + n^{3}

Or, in python:

def sum_of_cubes(n):
    return sum([i*i*i for i in range(n+1)])

These sums form a sequence 0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, ... and so there aren't many years which satisfy this equation. 2025 is the ninth in the sequence (if you start from 1, or tenth if you start from zero), so the previous year was 9³ years ago and the next one is 10³ years in the future — so this is probably the only one we'll live to see.

In summary, 2025 is the sum of the first 9 cubes. (if you start from 1)

Quartic in n

Many people will remember that the sum of the integers can be expressed more simply as a quadratic in n:

\sum_{i = 1}^{n} i = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}

This can be shown by pairing up the 1 with the n, and the 2 with the (n-1) and so on, creating n/2 pairs of (n+1), giving the total sum shown. Or you can show it a different way by considering the expansion of x² - (x-1)² and summing that from x up to n.

In a similar way, you can expand x⁴ - (x-1)⁴ and sum that, giving an expression for the sum of the cubes without the repeated addition:

C_{n} = \sum_{i = 1}^{n} i^{3} = 1^{3} + 2^{3} + 3^{3} + ... + n^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}

def quartic(n):
    return n**2 * (n+1)**2 // 4

for n = 9, $C_{9} = \frac{9^{2} 10^{2}}{4} Therefore,$ 2025 is also 9² * 5².

Square of the sum

You may have noticed that the formula for the sum of the cubes is similar to that of the single sum — in fact it is just the square:

C_{n} = \sum_{i = 1}^{n} i^{3} = {(\frac{n (n + 1)}{2})}^{2} = {(\sum_{i = 1}^{n} i)}^{2}

 def sum_squared(n):
    return sum([i for i in range(n+1)]) ** 2

We know from Kakuros and Sudokus that the sum of the first 9 digits is 45, so we can see that the square of this is 45² which is the same as 9² * 5².

Which means that 2025 is the square of the sum of the first 9 integers. (if you start from 1)

Summing the 9 times table

If you take each of the numbers from 1 to 9, and multiply them by each of the numbers from 1 to 9, you get a grid of 81 numbers from 1 to 81. Add these numbers up, and what do you get? 2025, of course!

Sum of the odd integers

Adding up the first k odd integers is the same as adding up (2i-1) instead of i. So we get:

1 + 3 + 5 + ... = \sum_{i = 1}^{k} 2 i - 1 = 2 \sum_{i = 1}^{k} i - \sum_{i = 1}^{k} 1

= 2 \frac{k (k + 1)}{2} - k = k^{2}

Which means that for our 45², 2025 is also the sum of the first 45 odd integers!

Or, in python:

def sum_of_odds(n):
    return sum([2*i+1 for i in range(n)])

Summary

So it seems that 2025 is indeed quite a special year. It's the sum of the first nine cubes, it's the product of 9-squared and 5-squared, it's the square of the sum of the first 9 integers, and it's the sum of the first 45 odd integers!

I hope you found some of this surprising and/or entertaining. There's more information at brilliant.org and numbersaplenty.com.

2025

Page last updated January 2025. Feedback welcome at mail@activityworkshop.net.