.... bs = ab+ab2 + ab3 + aba + ab3 + ........ ab”, and subtracting the above equation, there remains 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 be a geometrical progression of seven terms, of which the first = 3; 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 X 64192; and the whole progression will be 3, 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 72 If we multiply this last term 243 by the exponent, we shall have 2; the subtraction of the first term 4 leaves the remainder 5, which, divided by b-1, gives 5831. 665 461. 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 exponent, and the sums, so that s = 1 + } + { + } + ìì + 31⁄2 + o's + &c. ad infinitum. If we multiply by 2, we have 28 = 2 + + } + i +} +31 + &c. ad infinitum. 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 8 = 1 +} + ¿+&+¿+ &c. ad infinitum. Multiplying the whole by 3, we have 3s=3+1+1 + ↓ + 1⁄2 + &c. ad infinitum ; and subtracting the value of s, there remains 2 s = 3; wherefore the sum s 88 463. Let there be a progression, whose sums, first term = 2, and exponent; so that s = 2 + 3 + ÷ +33 +18+ &c. ad infinitum. Multiplying by, we have s = 3+2+3+?+3} + {1+ &c. ad infinitum. Subtracting now the progression s, there remains s; wherefore the sum required 464. If we suppose, in general, the first term = a, nent of the progression = b = 8. and the expo so that this fraction may be less than 1, and consequently c greater than b; the sum of the progression carried on, ad infinitum, will be found thus; And, subtracting this equation from the preceding, there remains If we multiply both terms of this fractions by c, we have s= ac с The sum of the infinite geometrical progression proposed is, there fore 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 this equation to the preceding, we obtain 24 466. We see, then, that if the first term a = 3, and the exponent, that is to say, b = 2 and c = 5, we shall find the sum of the progression ++++ &c. = 1; since, by subtracting the exponent from 1, there remains, and by dividing the first term by that remainder, the quotient is 1. 5 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, nổ tuổi trở tro troi ot&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 So this fraction, we have for the sum of the given progression. that taking only one term of the progression, namely, r, the error would be. 33 Taking two terms 1+1=1%, there would still be wanting To to make the sum = }}. 468. Another Example. Let there be given the infinite progression, 9+ 10+ 180 + 1000 + TO8oo + &c. The first term is 9, the exponent is. So that 1, minus the 용 exponent, = ; and 10, the sum required. 10 This series is expressed by a decimal fraction, thus 9,9999999, &c. CHAPTER XI. Of Infinite Decimal Fractions. 469. It will be very necessary to show how a vulgar fraction may be transformed into a decimal fraction; and, conversely, how we may express the value of a decimal fraction by a vulgar fraction. a 470. Let it be required, in general, to change the fraction, into a decimal fraction; as this fraction expresses the quotient of the division of the numerator a by the denominator b, let us write, instead of a, the quantity a,0000000, whose value does not at all differ from that of a, since it contains neither tenth parts, nor hundredth parts, &c. Let us now divide this quantity by the number b, according to the common rules of division, observing to put the point in the proper place, which separates the decimal and the integers. This is the whole operation, which we shall illustrate by some examples. Let there be given first the fraction, the division in decimals will assume this form, Hence it appears, that is equal to 0,5000000 or to 0,5; which is sufficiently evident, since this decimal fraction represents, which is equivalent to 1. 471. Let be the given fraction, and we have, This shows, that the decimal fraction, whose value is, cannot, strictly, ever be discontinued, and that it goes on ad infinitum, repeating always the number 3. And, for this reason, it has been already shown, that the fractions 3% + 180 + 1000 + 10800 &C. ad infinitum, added together make ΤΟ The decimal fraction, which expresses the value of, is also continued ad infinitum for we have, And besides, this is evident from what we have just said, because is the double of. 472. If be the fraction proposed, we have So that is equal to 0,2500000, or to 0,25; and this is evident, since+0 = 10% = 1. In like manner, we should have for the fraction, 75 So that 0,75; and in fact 7% +10=10% = 1. то The fraction is changed into a decimal fraction, by making When the denominator is 6, we find = 0,1666666, &c. which is. equal to 0,666666 0,5. Now 0,666666, and 0,5 = 1, wherefore 0,1666666 == } We find, also, = 0,333333, &c.; but becomes 0,5000000. Further, =0,833333 = 0,333333 + 0,5, that is to say, + 1 = /• |