I’ve delved deeply into divisiblity over the last few weeks, and am now examining properties of squares and sums of squares. I offer the following proof by induction (some algebraic manipulation is required as well).

It is known that the sum of the squares of the first *n *natural numbers is equal to

[( n )( n + 1 )( 2n + 1 )]/6

**PROOF BY INDUCTION**

Base cases: n = 1

[( 1 )( 1+1 )( 2(1)+1 )]/6 = [( 1 )( 2 )( 3 )]/6 = 6/6 = 1 = 1^{2}.

n = 2

[( 2 )( 2 + 1 )( 2(2) + 1 )]/6 = [( 2 )( 3 )( 5 )]/6 = 30/6 = 5 = 1^{2 }+ 2^{2}.

Induction Hypothesis

[( k )( k + 1 )( 2k + 1 )]/6 = ( **SUM** from i = 0 to i = k of i^{2} )

Induction Step

(k + 1)^{2 }+ ( **SUM** *from i = 0 to i = k of *i^{2} ) = ( **SUM** *from i = 0 to i = ***k + 1** of i^{2} )

**RHS**

( **SUM** *from i = 0 to i = ***k + 1** of i^{2} )Â = [( k + 1 )(( k + 1 ) + 1)( (2k + 1) + 1 )]/6

**LHS**

= ( k + 1 )^{2} + [( k )(k + 1 )( 2k + 1 )]/6

= [6( k + 1 )^{2 }+ ( k )(k + 1 )( 2k + 1 )]/6

= ( k + 1 )[6( k + 1 ) + k( 2k + 1 )]/6

= ( k + 1)[6k + 6 + 2k^{2} + k]/6

= (k + 1)[2k^{2} + 7k + 6]/6

= ( k + 1 )( k + 2 )( 2k + 3 )/6

= ( k + 1 ) (( k + 1) + 1)(( 2k + 1 ) + 1) /6 = **RHS**

**Q.E.D.**

I prefer proof by induction as it feels robust.

### Like this:

Like Loading...

## Published by E. E. Esterly

Hi! I'm Elizabeth, and I like to program robot brains, write software, think about and play around with numbers, and hang out with my awesome husband (also a programmer) and my dog (most certainly not a programmer). I'm a graduate student in Computer Science department at the University of New Mexico.
I came to Computer Science and Robotics from a non-traditional background. I'm a first-generation college student with a BFA in Fine Art and minors in French and Japanese. After receiving my art degree, I traveled the world with my husband tattooing for a few years before becoming increasingly interested in robotics and Computer Science...and the rest is history!
View all posts by E. E. Esterly