पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/२५१

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CHAPTER III--FRACTIONS.

three, the first and the last (denominators so obtained) being (however) multiplied (again) by 2 and ${\tfrac {2}{3}}$ (respectively).

Examples in illustration thereof.

76. The sum of five or six or seven (different fractional) quantities, having 1 for (each of) their numerators, is 1 (in each case). O you, who know arithmetic, say what the (required) denominators are.

The rule for finding out the denominators in the case of an even number (of fractions):--

77. When the sum of the (different fractional}quantities, having one for each of their numerators, is one, the (required) denominators are such as, beginning with two, go on (successively) rising in value by one, each (such denominator) being (further) multiplied by that

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as ${\tfrac {1}{2}}-{\tfrac {1}{\tfrac {2}{3}}}\times {\tfrac {1}{3^{n-1}}}$. From this it is clear that, when the first fraction ${\tfrac {1}{2}}$ and last fraction ${\tfrac {1}{{\tfrac {2}{3}}.3^{n-1}}}$ are added to this result, the sum becomes 1.

In this connection it may be noted that, in a series in goemetrical progression consisting of n terms, having ${\tfrac {1}{a}}$ as the first term and ${\tfrac {1}{a}}$ as the common ratio, the sum is, for all positive integral values of a, less than ${\tfrac {1}{a-1}}$ by ${\tfrac {1}{\tfrac {a-1}{a}}}\times$ the (n+1)th term in the series. Therefore, if we add to the sum of the series in goemetrical progression ${\tfrac {1}{\tfrac {a-1}{a}}}\times$ the (n + 1)th term which is the last fraction according to the rule stated in this stanza, we get ${\tfrac {1}{a-1}}$. To this ${\tfrac {1}{a-1}}$, we have to add ${\tfrac {a-2}{a-1}}$ in order to get 1 as the sum. This ${\tfrac {a-2}{a-1}}$ is mentioned in the rule as the first fraction, and so 3 is the value chosen for a, since the numerator of all the fractions has to be 1.

77. Here note ${\tfrac {1}{2\times 3\times {\tfrac {1}{2}}}}+{\tfrac {1}{3\times 4\times {\tfrac {1}{2}}}}+{\tfrac {1}{4\times 5\times {\tfrac {1}{2}}}}+....+{\tfrac {1}{(n-1)\times n\times {\tfrac {1}{2}}}}+{\tfrac {1}{n\times {\tfrac {1}{2}}}}$

== $2{\big \{}{\tfrac {1}{2\times 3}}+{\tfrac {1}{3\times 4}}+{\tfrac {1}{4\times 5}}+....+{\tfrac {1}{n(n-1)}}+{\tfrac {1}{n}}{\big \}}$

== $2{\big \{}\left({\tfrac {1}{2}}-{\tfrac {1}{3}}\right)+\left({\tfrac {1}{3}}-{\tfrac {1}{4}}\right)\left({\tfrac {1}{4}}-{\tfrac {1}{5}}\right)+....+\left({\tfrac {1}{n-1}}-{\tfrac {1}{n}}\right)+{\tfrac {1}{n}}{\big \}}$

== $2\times {\tfrac {1}{2}}==1.$