पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/२०९

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एतत् पृष्ठम् परिष्कृतम् अस्ति
13
CHAPTER II -- ARITHMETICAL OPERATIONS.

begin with 1 and end with 6, and then become gradually diminished, are divided between 441 persons. What is the share of each?

28. Gems (amounting to) 28483 (in number) are given (in offering) to 13 Jina temples. Give out the share of each (temple).

Thus ends division, the second of the operations known as Parikarman.

Squaring.

The rule of work in relation to the operation of squaring, which is the third (among the Parikarman operations), is as follows:--

29. The multiplication of two equal quantities: or the multiplication of the two quantities obtained (from the given quantity) by the subtraction (therefrom), and the addition (thereunto), of any chosen quantity, together with the addition of the square of that chosen quantity (to that product): or the sum of a series in arithmetical progression, of which 1 is the first term, 2 is the common difference, and the number of terms wherein is that (of which the square is) required : gives rise to the (required) square.

30. The square of numbers consisting of two or more places is (equal to) the sum of the squares of all the numbers (in all the places) combined with twice the product of those (numbers) taken (two at a time) in order

28. Here, 28483 is given as 83 + 400 + (4000 x 7).
25. The rule given herein, expressed algebraically, comes out thus :
(i) a x a =a2; (ii) (a+x)(a-x) + x2 = a2; (īi) 1+3+5+7+ . . . up to a terms == a2.
30. The word translated by place here is स्थान; it obviously means a a place in notation. Here, as a commentary interprets it, it may also denote the component parts of a sum, as each such part has a place in the sum. According to both these interpretations the rule works out correctly.
For instance, ${\displaystyle (1234)^{2}=(1000^{2}+200^{2}+30^{2}+4^{2})+2*\overbrace {1000*200} +2*\overbrace {1000*30} +2*\overbrace {1000*4} +2*\overbrace {200*30} +2*\overbrace {200*4} +2*\overbrace {30*4} .}$
Similarly ${\displaystyle (1+2+3+4)^{2}=(1^{2}+2^{2}+3^{2}+4^{2})+2(1*2+1*3+1*4+2*3+2*4+3*4)}$.