The sum of the squares of the first n natural numbers

The sum of the squares of the first n natural numbers

Nicholas Kampouras

Mechanical Engineer M.Sc. AMIMechE

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Abstract

Sequence is a group of items set in a specific order. In mathematics “sequence” usually means an ordered list of numbers. The natural numbers are the source of various well known and very interesting sequences. The sum of the terms of a sequence is called series. If a series is converging then the sum has a finite limit. If the series is diverging then such limit does not exist. Consider the sequence and the series of the squares of the natural numbers. It is diverging. That means that the sum of the terms or any partial sum has no finite limit. But what will happen if somebody wants to find the arithmetical value of the sum of the squares of the first thousand natural numbers? Does he have to calculate thousand squares and then adding up these values? Or there is a formula by which he can come up fast and easy with the desired result. The present article presents a method by which we can calculate the arithmetical value of the sum of the squares of the first n natural numbers.

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Consider the sequence 1, 2², 3², 4², …, n²

Rewriting the sequence it is  1², 2², 3², 4², …, n²

Visualizing this sequence we can say that we have the area of a 1x1 square then the area of a 2x2 square, the area of a 3x3 and so on (fig.1).

fig.1

Rewriting again the sequence we have 1x1², 1x2², 1x3², 1x4², …, 1xn², or in simple words we have a brick 1 unit long 1 unit wide and 1 unit tall, then a second brick 2 units long 2 units wide 1 unit tall, then a third one 3 units long 3 units wide 1 unit tall etc (fig.2)

fig.2

The series of the squares of the natural numbers can be written:

1 + 2²  + 3² + 4² +…+ n² = 1²  + 2² + 3² + 4² +…+ n² = (1x1²) + (1x2²) + (1x3²) + (1x4²) +  …+(1xn²)

In other words we have to add up the volumes of all the n bricks.

Consider the following example: Calculate the sum of the volumes of two bricks, brick A and brick B. The size of the first one is 2x2x1 and of the second is 3x3x1.

Let’s take the two bricks and stack the first on the top of the second, centered as in fig.3. All we have to do is to calculate the volume of the new solid which looks like a truncated pyramid.

Draw a line from each top corner of the 2x2x1 brick to the correspondent top corner of the 3x3x1 brick and extend them to the ground as in fig.3. The lines from the points E, F, G, and H intersect the ground on the points A, B, C and D. The parallelogram ABCD is a 4x4 square.

fig.3

The ABCDEFGH solid is a truncated pyramid.

We can split the pyramid in two halves, each one of them is a smaller truncated pyramid. The lower pyramid is ABCD1234 and the upper pyramid EFGH1234 fig.4. The volume of the lower pyramid consists of the volume of the brick plus the volumes of four equal sizes prisms and the volumes of eight equal sizes pyramids, the last ones located at the corners of the truncated pyramid fig.4.

Hence

V_{TR. PY.} = V_BRICK + 4V_PRISM + 8V_PYR.

fig.4

The base of the prism is a right angled triangle with a base of 1/2 and a height of 1. The length of the prism is equal to the side of the brick fig.5

The “corner” pyramids have a right angled triangle base 1/2 x1/2 and their height is 1 fig.5.

The same applies for the upper pyramid EFGH1234.

fig.5

:                                                                               
V_{TR. PY.} = V_BRICK + 4V_PRISM + 8V_PYR.

V_BRICK = V_{TR. PY.} – 4V_PRISM – 8V_PYR.

V_{BRICK =} (1/3)*H_TR.PY.(B+b+S(Bb)) – 4(1/2*0.5*1)*L_PRISM – 8(1/3*1/2*0.5*0.5*1)

V_{BRICK =} (1/3)*H_BRICK(B+b+S(Bb)) – L_BRICK – 1/3

B  is the area of the big base of the truncated pyramid and b is the area of the small base. The volumes of the “corner” pyramids are independent of the length “L” of the brick.
Then the sum of the volumes of the two bricks is:
The volume of the truncated pyramid ABCDEFGH (fig.4), minus the length of brick A, minus the length of brick B, minus two times 1/3.

V_A+V_{B  =} (1/3)(H_A+ H_B)(B+b+S(Bb)) – L_A – L_B – 2/3

V_A+V_{B  =} (1/3)(1+ 1)(2²+4²+S(4²x2²)) – 2 – 3 – 2/3

V_A+V_{B  =} (1/3)*2*(4+16+8) – 2 – 3 – 2/3 = 13

That means the method of the truncated pyramid for the calculation of the squares of the natural numbers 2 and 3 is correct because 2² + 3² = 13.
Now consider that we have ten bricks having the same height all of them H=1 and each one having a length and a width corresponding to the first ten natural numbers. We pile one on the top of the other as in fig.6. The area of the small base of the formed truncated pyramid is 1x1 and the area of the big base is 11x11. The height of the same pyramid is 10. The sum of the volumes of the ten bricks is:

fig.6

The area of the big base of the truncated pyramid is B =11x11=121 and the area of the small base b =1x1=1.

V₁+V₂+V₃+ V₄+V₅+V₆+ V₇+V₈+V₉+V₁₀ =

 = (1/3)(H₁+H₂+H₃+H₄+H₅+H₆+H₇+H₈+H₉+H₁₀)(B+b+S(Bb)) –

–       (L₁ +L₂+ L₃ +L₄+ L₅ +L₆+ L₇ +L₈+ L₉ +L₁₀) – (10/3) =>

V₁+V₂+V₃+ V₄+V₅+V₆+ V₇+V₈+V₉+V₁₀ =

= (1/3)(1+1+1+1+1+1+1+1+1+1)(11²+1²+S(11² 1² )) –

(1 +2+ 3 +4+ 5 +6+ 7 +8+ 9 +10) – (10/3) = 385 

The result is correct because 1² + 2² + 3² + 4² + 5² + 6² + 7² + 8² + 9² + 10² = 1+4+9+16+25+36+49+64+81+100 = 385.

Now if we consider n bricks where n = 1,2,…,n the truncated pyramid which is formed has a small base with an area of b = 1x1 = 1 and a big base with an area of B = (n+1)(n+1) = (n+1)². The total volume is:

V₁+V₂+V₃+…+V_n =

= (1/3)(H₁+H₂+H₃+…+H_n)(B+b+S(Bb)) –

– (L₁ +L₂+ L₃ +…+L_n) – (n/3) =>

V₁+V₂+V₃+…+V_n =

= (1/3)*(1+1+1+…+1)((n+1)²+1²+(n+1)) –

– (1 +2+ 3 +…+n) – (n/3) =

= (n/3)*((n+1)²+1²+(n+1)) – (1 +2+ 3 +…+n) – (n/3) =

= (n/3)*((n+1)²+n+1) – (1 +2+ 3 +…+n) =

= (n/3)*(n²+3n+2) – (1 +2+ 3 +…+n) =

= (n³/3) + n² + (2n/3) – (1 +2+ 3 +…+n)

 The sum of the first n natural numbers is:

1 + 2 + 3 +…+ n = ((n+1)n)/2 thus

V₁+V₂+V₃+…+V_n = 1*1² + 1*2² +1*3² +…+ 1*n² =

1² + 2² +3² +…+ n² = n³/3 + n² + (2n/3) – n²/2 – n/2 =>

1² + 2² +3² +…+ n² = (n³/3) + (n²/2) + (n/6)