found, by multiplying the n- 1 power of b, or 6-1, 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 =249. Now 2° being = 512; 210 will be = 1024. Wherefore the square of 210, or 220 1048576, and the square of this number, or 1099511627776 = 240. Multiplying therefore this value of 24° by 2°, or by 512, we have 249 equal to 562949953421312. H 452. 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 2 s = 2 + 4 + 8 + 16 + 32 +64 + 128 +256 + 5121024. Subtracting from this the progression represented by s, there remains s = 1024 11023; wherefore the sum required is 1023. 453. Suppose now, in the same progression, that the number of terms is undetermined and = n, so that the sum in question, or s, = 1 + 2 + 2a + 2a + 2a . . . . 2′′−1. If we multiply by 2, we have 282 + 2a + 23 + 24.... 2", and subtracting from We see, this equation the preceding one, we have s = 2′′ — 1. therefore, that the sum required is found, by multiplying the last term, 2"-1, by the exponent 2, in order to have 2", and subtracting unity from that product. 454. This is made still more evident by the following examples, in which we substitute successively, for n, the numbers 1, 2, 3, 4, &c. 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=63, &c. 455. On this subject the following question is generally proposed. A man offers to sell his horse by the nails in his shoes, which are in number 32; he demands 1 liard for the first nail, 2 for the second, 4 for the third, 8 for the fourth, and so on, demanding for each nail twice the price of the preceding. It is required to find what would be the price of the horse? = This question is evidently reduced to finding the sum of all the terms of the geometrical progression 1, 2, 4, 8, 16, &c. continued to the 32d term. Now this last term is 231; and, as we have already found 220 = 1048576, and 210=1024, we shall have 920 x 210 230 equal to 1073741824; and multiplying again by 2, the last term 2312147483648; doubling therefore this number, and subtracing unity from the product, the sum required becomes 4294967295 liards. These liards make 1073741823 sous, and dividing by 20, we have 53687091 livres, 3 sous, 9 deniers for the sum required. = 456. Let the exponent 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. Suppose its, so that S=1+3+9+27+81 +243 + 729; we shall then have, multiplying by 3, 3 s = 3 +9 + 27 + 81 + 243 + 729 + 2:87; and subtracing the preceding series, we have 2 s 2187-1= 2186. So that the double of the sum is 2186, and consequently the sum required = 1093. ... · 457. In the same progression, let the number of terms = n, and the sums; so that s = 1 + 3 + 3o + 33 +5* + Sn-1. If we multiply by 3, we have 3 s = 3 + 32 + 33 + 3a + 3". Subtracting from this the value of s, as all the terms of it, except the first, destroy all the terms of the value of 3 s, except the last, we shall have 2 s = 3" 1; therefore s= 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. This will appear, also, from the following examples; 3" -1 2 458. Let us now suppose, generally, the first term = a, the exponent=b, the number of terms = n, and their sum = s, so that s=aab+ab2 + ab3 + a ba +.... a bn-1. If we multiply by b, we have bsabab2 + ab3 + aba +abs +.... ab", and subtracting the above equation, there remains (b whence we easily deduce the sum required s= 1) s = a bn a a; Conse quently, the sum of any geometrical progression is found by multiplying the last term by the exponent of the progression, subtracting the first term from the product, and dividing the remainder by the exponent minus unity. 459. Let there by a geometrical progression of seven terms, of which the first = 5; and let the exponent be = 2; we shall then have a = 3, b = 2, and n = 7; wherefore the last term = 3 × 2o, or 3 × 64 = 192; and the whole progression will be S, 6, 12, 24, 48, 96, 192. Further, if we multiply the last term 192 by the exponent 2, we have 384; subtracting the first term there remains 381; and dividing this by b— 1, or by 1, we have 381 for the sum of the whole progression. 460. Again, let there be a geometrical progression of six terms; let 4 be the first, and let the exponent be. The progression is 4, 6, 9, 27, 1,243. If we multiply this last term 243 by the exponent, we shall have 2; the subtraction of the first term 4 leaves the remain-der 665, which divided by b = 1, gives 85 = 831 6 6 461. When the exponent is less than 1, and consequently, when the terms of the progression continually diminish, the sum of such a decreasing progression, which would go on to infinity, may be accurately expressed. For example, let the first term = 1, the exporent = 1, and the sums, so that s = 1 + 2 + 3 + ÷ + 8 +3 +64 + &c. ad infinitum. If we multiply by 2, we have ad infinitum. 2 s = 2 + + + + + 8 ++ &c. And, subtracting the preceding progression, there remains s=2 for the sum of the proposed infinite progression. 462. If the first term = 1, the exponent, and the sum =s; so that s = 1 + } + 3 + 37 +31 + &c. ad infinitum. Multiplying the whole by 3, we have 3s=3+1+} + { + 7 + &c. ad infinitum ; and subtracting the value of s, there remains 2 s3; wherefore the sum s = 1. 81 465. Let there be a progression, whose sums, first term =2, and exponent; so that s = 2+++ 3 3 +12+ &c. ad infinitum. = 81 Multiplying by, we have s÷+2+} +÷+11 + 1/2 728 + &c. ad infinitum. Subtracting now the progression s, there remains s; wherefore the sum required = 8. 464. If we suppose, in general, the first term = a, and the less than 1, and consequently c greater than b; the sum of the progression carried on, ad infinitum, will be found thus ; ab a b2 Makes a + + Multiplying by C cc a b3 + a b4 C4 + &c. + 3+ a bз abs C4 b ab + &c. ad infinitum. And, subtracting this equation from the preceding, there re If we multiply both terms of this fraction by c, we have 8= ac c-b The sum of the infinite geometrical progression proposed is, therefore, found by dividing the first term a by 1 minus the exponent, or by multiplying the first term a by the denominator of the exponent, and dividing the product by the same denominator diminished by the numerator of the exponent. 465. In the same manner, we find the sums of progressions, the terms of which are alternately affected by the signs + and Let for example, And, adding to this equation to the preceding, we obtain (1 +-) s a = α. Whence we deduce the sum required s= ac c+b⋅ 6 24 466. We see, then, that if the first term a = }, and the exponent, that is to say, b = 2 and c= 5, we shall find the sum of the progression 3+2+13 +332327+&c. = 1; since, by subtracting the exponent from 1, there remains 3, and by dividing the first term by that remainder, the quotient is 1. Further, it is evident, if the terms be alternately positive and negative, and the progression assume this form; 467. Another example. Let there be proposed the infinite progression, TO +180 + 7000 + 70800 + 100000 + &c. The first term is here, and the exponent is. Subtracting this last from 1, there remains, and, if we divide the first term by this fraction, we have for the sum of the given progression. So that taking only one term of the progression, namely, the error would be. |