Sum of Powers of Numbers - A Derivation of Bernoulli/Faulhaber Formula

Sum of Powers of Natural Numbers - Exciting to Explore Different Ways of Doing It

'While there is one correct answer, there can be multiple ways to solve a problem, encouraging creativity and different perspectives' 

I have always found numbers fascinating - they are versatile, form patterns, help develop creative thinking and more. Satisfaction of solving a problem is a wonderful feeling, greatly enhanced by finding yet another way to do so. 

I remember that one of the first problems at school was summing the first N natural numbers -  the answer S = N(N+1)/2 was amazing; you can set N = 100 and instantly get the sum of all the numbers from 1 to 100.  

In this blog, I wish to discuss several different ways of summing the powers of natural numbers.  The motivation to write this blog came from the way many people reacted to my previous blog about the amazing properties of the number 2025 - in particular the fact that the square of the sum of numbers from 1 to 9 is equal to the sum of their cubes. 

      2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 +9)^2 

        2025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 

I can fully understand the confused looks as the result is so counterintuitive, yet true.  Of course, Wiki has a way to make it look obvious as shown in the next slide:  

The figure shows the result for N = 5, but it is easy to peal off squares or add more to demonstrate it for other value of N.  

Sum of Natural Numbers:   First, I shall discuss the sum of first power of natural numbers. We can write the sum in two different ways:

S = 1 + 2 + 3 +...................+ (N -1) + N  ...    eq.1   (N terms here)

S = N + (N -1) + ..................+ 2 + 1       ... eq. 2   (N terms)

Adding the two equations will give us

2S = (N+1) + (N+1) + ..........(N+1) + (N+1)    (N terms)

Or     2S = N(N+1)   giving us   S = N(N+1)/2     ... eq.3

I find it fascinating how maths works:  Multiply eq.1 by 2 to obtain

2S = 2 + 4 + 6 + ...................+ 2N = N(N+1)    ... eq.4  (N terms)

Subtract 1 from each digit in the above equation

      1 + 3 + 5 + ................+(2N -1) = N(N+1) - N    ... eq.5

The LHS has N terms - that is why we subtract N on the RHS

Hence,    1 + 3 + 5 + ....  +(2N -1) = N^2 + N - N = N^2     ...  eq.6

What eq.6 says is that the sum of the first N odd numbers is equal to N^2.   

Eq.4 tells us that  the sum of the first N even numbers is N(N+1) = N^2+N.

I shall now discuss various ways by which we can show that the square of the sum of numbers 1 to N is equal to the sum of cubes of numbers from 1 to N.  We shall also look at the sum of higher powers of natural numbers.

Final Word: We have discussed several methods of summing powers of natural numbers.  The methods have great heuristic value and may be used in a classroom situation to encourage students.

However, the best situation is if the sum of powers may be expressed as a closed expression for all powers - it would then be a matter of working out the expression for a chosen power quotient.  Such an expression is available and is called the Faulhaber's formula.  The derivation of the Faulhaber's formula generally requires advanced knowledge of mathematics.  In Part 2, I shall use a generating function approach to derive the Faulhaber's formula - a method that is tractable with higher school level mathematics.