Using sum of squares triangle to prove sum of squares formula

SumOfSquare

We construct a number triangle from the integers 1, 2, 3, . . . n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2. . . The last column contains only one integer n. Here is an example with n = 5.

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1

Now we calculate T(n), the sum of all numbers in the triangle by two ways: By row and by column.

1. Calculation by column:

T(n) =

1*(2n-1) + 2*(2n-1-2) + 3*(2n-1-4) + . . . + i*(2n-1-2*(i-1)) + . . . + n*(2n-1-2(n-1))

1*(2n+1-2) + 2*(2n+1-4)+ 3*(2n+1-6) + . . . + i*(2n+1-2*i) + . . . + n*(2n+1-2n)

(2n+1)*(1 + 2 + . . . + n) – 2*(1^2 + 2^2 + . . . + n^2)

(2n+1)*(n+1)*n/2 – 2*(1^2 + 2^2 + . . . + n^2)

So T(n) = (2n+1)*(n+1)*n/2 – 2*(1^2 + 2^2 + . . . + n^2)

T(n) = (2n+1)*(n+1)*n/2 – 2*S(n) where S(n) = (1^2 + 2^2 + . . . + n^2) (1)

2. Calculation by row:

Please note that the triangle is symmetrical by middle row 1, 2, 3 . . . n

T(n) =

2*(1 + (1+2) + (1+2+3) + . . . + (1+2+ . . . + n-1)) + (1+2+ . . . + n)

2*(1*(1+1)/2 + 2*(2+1)/2 + (3*(3+1)/2) + . . . + (n-1)*n/2) + n*(n+1)/2

1*(1+1) + 2*(2+1) + 3*(3+1) + . . . + (n-1)*(n-1+1) + n*(n+1)/2

1^1 +1 + 2^2 +2 + . . . + (n-1)^2 + (n-1) + n*(n+1)/2

1^1 + 2^2 + . . . + (n-1)^2 + 1 + 2 + . . . + (n-1) + n*(n+1)/2

1^1 + 2^2 + . . . + (n-1)^2 + n^2 + 1 + 2 + . . . + (n-1) + n*(n+1)/2 – n^2

S(n) + (n-1)*n/2 + n*(n+1)/2 – n^2

T(n) = S(n) + (n-1)*n/2 + n*(n+1)/2 – n^2 = S(n) (2)

From (1), (2)

(2n+1)*(n+1)*n/2 – 2*S(n) = S(n)

S(n) = (2n+1)*(n+1)*n/6

It is sum of square formula.

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About hinhoc

To discover new interesting: first observe all strange, abnormal but then make them familiar, normal!
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