Elements of Algebra

166 ELEMENTS SECT. III. examples, in which we substitute successively for $n$, the numbers, 1, 2, 3, 4, \ldots $1 = 1$; $1 + 2 = 3$; $1 + 2 + 4 = 7$; $1 + 2 + 4 + 8 = 15$; $1 + 2 + 4 + 8 + 16 = 31$; $1 + 2 + 4 + 8 + 16 + 32 = 32 \times 2 - 1 = 63$. 511. On this subject, the following question is generally proposed. A man offers to sell his horse on the following condition; that is, he demands 1 penny for the first nail, 2 for the second, 4 for the third, 8 for the fourth, and so on, doubling the price of each succeeding nail. It is required to find the price of the horse, the nails being 32 in number? This question is evidently reduced to finding the sum of all the terms of the geometrical progression 1, 2, 4, 8, 16, \ldots continued to the 32d term. Now, that last term is $2^{31}$; and, as we have already found $2^{20} = 1048576$, and $2^{10} = 1024$, we shall have $2^{20} \times 2^{10} = 2^{30} = 1073741824$; and multiplying again by 2, the last term $2^{31} = 2147483648$; doubling therefore this number, and subtracting unity from the product, the sum required becomes 4294967295 pence; which being reduced, we have 17895697l. 1s. 3d. for the price of the horse. 512. Let the ratio now be 3, and let it be required to find the sum of the geometrical progression 1, 3, 9, 27, 81, 243, 729, consisting of 7 terms. Calling the sum $s$ as before, we have $s = 1 + 3 + 9 + 27 + 81 + 243 + 729$. And multiplying by 3, $3s = 3 + 9 + 27 + 81 + 243 + 729 + 2187$. Then subtracting the former series from the latter, we have $2s = 2187 - 1 = 2186$: so that the double of the sum is 2186, and consequently the sum required is 1093. 513. In the same progression, let the number of terms be $n$, and the sum $s$; so that $s = 1 + 3 + 3^2 + 3^3 + 3^4 + \ldots 3^{n-1}$. If now we multiply by 3, we have $3s = 3 + 3^2 + 3^3 + 3^4 + \ldots 3^n$. Then subtracting from this expression the value of $s$, as before, we shall have $2s = 3^n - 1$; therefore $s = \frac{3^n - 1}{2}$. So that the sum required is found by multiplying the last term by 3, subtracting 1 from the product, and dividing the remainder by 2; as will appear, also, from the following particular cases: