Generalize the result:

[ sum cdots sumsum r = frac{ncdot(n+1)cdot(n+2)cdot dots cdot (n+k)}{1cdot 2 cdot 3 dots cdot (k+1)} ]

Chapter 4 is the largest chapter in the book. It contains 149 rules and 94 examples of geometric problems, including a number of approximation formulas for circular shapes. It should be noted a formula newly developed by Narayana for determining the area of a quadrilateral with the help of the so-called third diameter.

Quadruple ABCD results from alternating sides *NS* And *NS* Hypotenuse square ABED with diagonals *F* And *g*. The quadrant regions agree, since according to Brahmagupta’s formula, the region depends only on the length of the four sides of the quadrant: (s = frac{1}{2}cdot (a + b + c + d)). The area of the square ABED can be calculated as the sum of the areas of the triangle ABE with the sides *a, c, g* And the triangle dirhams with sides *b, d, g*:

[A_{ABED} = A_{ABE} + A_{AED} = frac{acdot c cdot g}{4R} + frac{g cdot b cdot d}{4R} = frac{g}{4R} (ac + bd)]

On the other hand, according to Ptolemy’s theorem, the product of the lengths of the diagonals in a hypotenuse quadrilateral is equal to the sum of the product of the opposite sides of the quadrilateral, i.e. *ac + bd = his*. Hence the following:

[A_{ABCD}= A_{ABED}= frac{g}{4R} (ac+bd) = frac{efg}{4R}]

One task deals with special triangles whose side lengths are natural numbers and which differ by only one unit; The height length on the side of the base should also be a natural number. Narayana realizes that the left part of the base has a length of 0.5*x* – It should contain 2, the right section accordingly 0.5*x* + 2, because according to the Pythagorean theorem, the following applies to two partial triangles:

[(x-1)^2 – (frac{1}{2}x-2)^2 = y^2 = (x+1)^2 – (frac{1}{2}x+2)^2]

This means: (y^2 = frac{3}{4}x^2 -3.) This equation has an infinite number of solutions: (4;3), (14;12), (52; 45 ), (194; 168), (724; 627), …

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