arrived at is) multiplied by the half of the given number. This gives rise to the combined sum of the square (of the given number), the cube (of the given number), and the sum of the natural numbers (up to the given number). This combined sum, divided by three, gives rise to the sum of the squares (of the given number of natural numbers).
Examples in illustration thereof.
297. (In a number of series of natural numbers), the number of natural numbers is (in order) 8, 18, 20, 60, 81, and 36. Tell me quickly (in each case) the combined sum of the square (of the given number), the cube (of the given number), and the sum of the given number of natural numbers. (Tell me) also the sum of the squares of the natural numbers (up to the given number).
The rule for arriving at the sum of the squares of a number of terms in arithmetical progression, whereof the first term, the common difference, and the number of terms are given:-
298.[1] Twice the number of terms is diminished by one, and (then) multiplied by the square of the common difference, and is (then) divided by six. (To this), the product of the first term and the common difference is added. The resulting sum is multiplied by the number of terms as diminished by one(To the product so arrived at), the square of the first term is added. This sum multiplied by the number of terms becomes the sum of the squares of the terms in the given series.
Again, another rule for arriving at the sum of the squares of a number of terms in arithmetical progression, whereof the first term, the common difference, and the number of terms are given:-
299. Twice the number of forms (in the series) is diminished by one, and (then) multiplied by the square of the common difference, and (also) by the number of terms as diminished by one. This
298.^ sum of the squares of the terms in a series in arithmetical progression.
![{\displaystyle \left[\left\{{\tfrac {(2n-1)^{2}}{6}}+ab\right\}(n-1)+a^{2}\right]n=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/273acae27f73155b8c84acf56c316939e70b11cc)
