[This particular series is a binomial series, the successive terms being the coefficients of x, x2, &c., in the expansion of (1-x)-3⁄4. Hence 1+S,= sum of the first (n+1) coefficients in the expansion of (1 − x)—3 = coefficients of x" in (1 − x) ̄† × (1 − x)−1, that is in (1 − x) ̄§]. Ex. 2. Find the sum of n terms of the series + Ans. 2 2.6 2.6.10 +... -1. + 1.2 1.2. where a is any integral expression of the rth degree in n, can be found in the following manner. - Hence S× (1-x)”+1 = a + {α, − (r + 1) a} x +..... Now a, is by supposition an integral expression of the rth degree in p; hence a1 = A‚p” + A‚-1 p1 + Ar-2 p2 + ... + A。 —2 where A, A,...A, do not contain p. Also, by Art. 301, the sum of the series is zero for all integral values of k less than r+1. Hence is zero for all values of p. All the terms of the product S, × (1-x) will there fore vanish except those near the beginning, or the end, for which the series a,- (r+1) ap-1+... is not continued for (r+ 2) terms, that is all the terms of the product will vanish except the first r+1 terms and the last r+1 terms. Hence S2 × (1 − x)+1 = α + {α, − (r + 1) α }x+..... n + {αr-1 − (r + 1) αr-2 + ... + ( − 1)” (r +1) α ̧} x′′ a。} .. (1-x)2 S2=1+xn+1 {(n − 2) (n + 1)} + (n + 1) xn+2, [all the other terms vanishing on account of the identity Ex. 2. Find the sum of n+1 terms of the series 13+23x+33x2+ . (n+1)31⁄23. ...... ..SX (1-x)=1+(23-4)x+(33-4. 23+6.13) x2 + (43 − 4 . 33 +6. 23 – 4. 13) x3 k3 − 4 (k − 1)3 + 6 (k − 2)3 – 4 (k − 3)3 + (k − 4)3 = 0 identically.] Hence Sn=[1 + 4x + x2 -- (n3 + 6n2+12n+8) x2+1 +(3n3 +15n2+21n+5) x2+2 - (3n3 +12n2 + 12n+4) x2+3 + (n+1)3 xn+4] / (1 − x)a. When x is numerically less than 1, the series is convergent, and the sum of the series continued to infinity is (1 + 4x + x2) / (1 − x)1. 323. Series whose law is not given. We have hitherto considered series in which the general term was given, or in which the law of the series was obvious on inspection. We proceed to consider cases in which the law of the series is not given. With reference to series in which the law is not given, but only a certain number of the terms of the series, it is of importance to remark that in no case can the actual law of the series be really determined: all that can be done is to find the simplest law the few terms which are given will obey. There are for instance an indefinite number of series whose first few terms are given by x+x2+x3+ ..., the simplest of all the series being the geometrical progression whose nth term is a": another series which has the given. terms is that formed by the expansion of " + x2 + x2 + ··· + x® x 1 x10 which agrees with the geometrical progression except at every 10th term. Note. In what follows it must be understood that by the law of a series is meant the simplest law which satisfies the given conditions. METHOD OF DIFFERENCES. 324. If in any arithmetical series a1 + a2+ α z +...+am each term be taken from the succeeding term, a new series is formed, namely the series (α2 − α ̧) + (α ̧—α2) +...+ (ɑn — αn-1) +........., which is called the first order of differences. If the new series be operated upon in the same way, the series obtained is called the second order of differAnd so forth. ences. Thus, for the series 2, 7, 15, 26, 40,..., the first order of differences is 5, 8, 11, 14,..., 325. When the law of a series is not given, it can often be found by forming the series of successive orders of differences; if the law of one of these orders of differences can be seen by inspection, the law of the preceding order of differences can often be found, and then the law of the next preceding order of differences, and so on until the law of the series itself is obtained. The method will be seen from the following examples. Ex. 1. Find the nth term of the series 1+6+23+58 +117 +206+ The first order of differences is 5+17+35+59 +89+.. The second order of differences is clearly an arithmetical progression whose nth term is 6 (n+1). Hence, if v be the nth term of the first order of differences, we vn have in succession Also v1=6.1-1. Hence, by addition, vn=6(1+2+...... + n) − 1 = 3n (n + 1) − 1. Then again, we have in succession un-un-1=Un-1=3 (n − 1) n−1; Un-1-un-2=3 (n − 2) (n − 1) −1; ... ; .. ; U2- U1 =3.1.2-1. Also u1 = 1. Hence un=3{(n−1)n+......1.2}-n+2=(n-1)n(n+1)−n+2. Ex. 2. Find the nth term and the sum of n terms of the series 6+9+14+23+ 40+..... The first order of differences is 3+5+9+17+..... Hence the second order of differences is a geometrical progression, the (n - 1)th term being 2"-1. Hence, if v be the nth term of the first order of differences, we have in succession Vn− Un-1=2n−1, Un−1− Un−2=2′′−2, Also v1=3. Hence, by addition, vn=(2+22+ = ..2n−1)+3=2n+ 1. U2-u1=21+1. Also u1 = 6. Then again, we have in succession un-un-1=vn-1=2′′−1+1, Hence un=(2n-1+...2)+n+5=2"+n+3. The sum of n terms of the series can now be written down: for the sum of n terms of the series whose general term is 2"+n+3 is (2+22+ 1 +2n) + {n + (n − 1) + ... + 1} + 3n=2n+1 − 2 + 2 n (n + 1) + 3n. Note. By the method adopted in the preceding examples the nth term of a series can always be found provided the terms of one of its orders of differences are all the same, or are in geometrical progression. 326. It is of importance to notice that when the nth term of a series is an integral expression of the rth degree in n, all the terms of the rth order of differences will be the same. |