CHAPTER V1-MIXED PROBLEMS. - 171 product is divided by six. (To this resulting quotient), the square of the first term and the (continued product) of the number of terms as diminished by one, the first term, and the common difference, are added. The whole (of this) multiplied by the number of terms becomes the required result. Examples in illustration thereof. 300. (In a series in arithmetical progression), the first term is 3, the common difference is 5, the number of terms is 5. Give out the sum of the squares (of the terms) in the series. (Similarly, in another series), 5 is the first term, 3 the common difference, and 7 the number of terms. What is the sum of the squares (of the terms) in this series? The rule for arriving at the sum of the cubes (of a given number of natural numbers) :- 301. The quantity represented by the square of half the (given) number of terms is multiplied by the square of the sun of one and the number of terms In this (science of) arithmetic, this result is said to be the sum of the cubes (of the given number of natural numbers) by those who know the secret of calculation. Examples an illustration thereof.. 302. Give out (in each case) the sum of the cubes of (the natural numbers up to) 6, 8, 7, 25 and 258. The rule for arriving at the sum of the cubes (of the terms in a series in arithmetical progression), the first term, the common difference, and the number of terms whereof are optionally chosen :- 303. The sum (of the simple terms in the given series), as multiplied by the first term (therein), is (further) multiplied by the 2 301. Algebraically, (2) *(n + 1)² = 83, which is the sum of the cubes of the natural numbers up to n. 303. Algebrancally, sa (ab) + sb the sum of the cubes of the terms in a series in arithmetical progression, where s= the sum of the simple terms of the series. The sign of the first tern in the formula is + or according ar a> or <b.
