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All numbers from 3 upwards in order are polygonal, containing as many angles as they have units, e.g. 3, 4, 5, etc.

"As with regard to squares it is obvious that they are such because they arise from the multiplication of a number into itself, so it was found that any polygonal multiplied into a certain number depending on the number of its angles, with the addition to the product of a certain square also depending on the number of the angles, turned out to be a square. This I shall prove, first showing how any assigned polygonal number may be found from a given side, and the side from a given polygonal number. I shall begin by proving the preliminary propositions which are required for the purpose."

I. If there are three numbers with a common difference, then 8 times the product of the greatest and middle + the square of the least a square, the side of which is the sum of the greatest and twice the middle number.

Let the numbers be AB, BC, BD in the figure, and we have to prove 8AB. BC + BD2 = (AB+ 2BC)2.

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By hypothesis AC= CD, AB=BC+CD, BD=BC-CD.
Now 8AB. BC= 4AB. BC + (4BC2 + 4BC. CD).

Therefore 8AB.BC+BD'

=4AB. BC+4BC2 + (4BC. CD + BD2)

= 4AB. BC + 4BC2+AB2,

[Eucl. II. 8]

and we have to see how AB+4AB. BC+4BC* can

be made a square.

[Diophantus does this by producing BA to E, so that AE= BC, and then proving that

AB2 + 4AB. BC+ 4BC2 = (BE+EA)2.]

It is indeed obvious that

AB2 + 4AB. BC + 4BC2 = (AB + 2BC)2.

2. If there are any numbers, as many as we please, in A.P., the difference between the greatest and the least is equal to the common difference multiplied by the number of terms less one.

[That is, if in an A.P. the first term is a, the common difference b and the greatest term 1, n being the number of terms, then

l_a = (n - 1) b.]

Let AB, BC, BD, BE have a common difference.

A

Now AC, CD, DE are all equal.

B

Therefore EA = AC× (number of terms AC, CD, DE) = ACX (number of terms in series - 1).

3. If there are as many numbers as we please in A.P., then (greatest + least) x number of terms double the sum of the

terms.

[That is, with the usual notation, 2s = n (l+a).]

(1) Let the numbers be A, B, C, D, E, F, the number of them being even.

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Let GH contain as many units as there are numbers, and let GH, being even, be bisected at K. Divide GK into units at L, M.

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(2) Let the number of terms be odd, the terms being

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Let there be as many units in FG as there are terms,

so that there is an odd number of units.

Let FH be one of them; bisect HG at K, and divide HK

into units, at L.

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4 If there are as many numbers as we please beginning with I and increasing by a common difference, then the sum of all × 8 times the common difference + the square of (common difference 2) = a square, the side of which diminished by 2 = the common difference multiplied by a number which when increased by 1 is double of the number of terms.

=

[The A.P. being 1, 1 + b, ... I + (n − 1)b, and s the sum, we have to prove that

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The proof being cumbrous, I shall add the generalised algebraic equivalent in a column parallel to the text.]

Let AB, CD, EF be the terms in

A.P. after I.

in |

1 + b, 1 + 26, 1+ 3b,....

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= 4n (n − 1) b2

Is then 4GH.HM.KB +4(GH+HM) KB+KB2+KN2 | + 4 {n + (n − 1)} b + b2 + 22

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[5] If there be as many terms as we please in A.P. beginning from 1, the sum of the terms is polygonal; for it has as many angles as the common difference increased by 2 contains units, and its side is the number of the terms set out including 1."

The numbers being as set out in the figure of Prop. 4, we have, by that proposition,

(sum of terms). 8KB + NB2 = KƠ.

Taking another unit AP, we have KP=2, while KN=2;
therefore PB, BK, BN are in A.P., so that
8PB.BK + NB = (PB + 2KB)2;

[Prop. 1]

and PB+2KB-2= PB+2KB-PK = 3KB,

while 3+1 = 2. 2, or 3 is one less than the double of 2. Now, since the sum of the terms of the progression

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Therefore HF is a mean proportional between the two squares,

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