Elements of Algebra

CHAP. XI. OF ALGEBRA. 165 $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}$, &c. 507. Here, therefore, we have to consider: 1. The first term, which we have called $a$. 2. The exponent, which we call $b$. 3. The number of terms, which we have expressed by $n$. 4. And the last term, which, we have already seen, is $ab^{n-1}$. So that, when the first three of these are given, the last term is found by multiplying the $n-1$ power of $b$, or $b^{n-1}$, by the first term $a$. If, therefore, the 50th term of the geometrical progression $1, 2, 4, 8$, &c. were required, we should have $a=1$, $b=2$, and $n=50$; consequently the 50th term would be $2^{49}$; and as $2^9 = 512$, we shall have $2^{10} = 1024$; wherefore the square of $2^{10}$, or $2^{20}$, $= 1048576$, and the square of this number, which is $1099511627776$, $= 2^{40}$. Multiplying therefore this value of $2^{40}$ by $2^9$, or $512$, we have $2^{49} = 562949953421312$ for the 50th term. 508. One of the principal questions which occurs on this subject, is to find the sum of all the terms of a geometrical progression; we shall therefore explain the method of doing this. Let there be given, first, the following progression, consisting of ten terms: $1, 2, 4, 8, 16, 32, 64, 128, 256, 512$, the sum of which we shall represent by $s$, so that $s=1+2+4+8+16+32+64+128+256+512$; doubling both sides, we shall have $2s=2+4+8+16+32+64+128+256+512+1024$; and subtracting from this the progression represented by $s$, there remains $s = 1024 - 1 = 1023$; wherefore the sum required is $1023$. 509. Suppose now, in the same progression, that the number of terms is undetermined, that is, let them be generally represented by $n$, so that the sum in question, or $s, = 1+2+2^2+2^3+2^4 \dots 2^{n-1}$. If we multiply by $2$, we have $2s=2+2^2+2^3+2^4+2^5 \dots 2^n$; then subtracting from this equation the preceding one, we have $s = 2^n - 1$; or, generally, $s = 2^n - 1$. It is evident, therefore, that the sum required is found, by multiplying the last term, $2^{n-1}$, by the exponent $2$, in order to have $2^n$, and subtracting unity from that product. 510. This is made still more evident by the following