THE DIFFERENTIAL METHOD OF SERIES.

ART. 324. A Series consists of a number of terms, each of which is derived from one or more of the preceding terms, according to some determinate law. (Art. 134.)

The use of the differential method is, 1st, to find the successive differences of the terms of a series; 2nd, any particular term of the series; or, 3rd, the sum of a finite number of its terms.

If, in any series, we take the first term from the second, the second from the third, the third from the fourth, and so on, the new series thus formed is called the First order of differences.

If we proceed with this new series in the same manner, we shall obtain another series termed the Second order of differences. In a similar manner we find the third, fourth, &c., orders of differences.

Thus, if we have the series

1, 8 27, 64, 125, 216, ..

the 1st order of differences is 7, 19, 37, 61, 91

ART. 325. PROBLEM 1.- To find the first term of any order of

then the respective orders of differences are,

1st order, b-a c-b

d-c e-d,

.

2nd order, c-2b+a, d-2c+b, e—2d+c,

Here each difference pointed off by commas, though a compound quantity, is called a term. Thus the first term in the 1st order is b-a; in the second order c-2b+a, &c.

If we denote the first terms in the 1st, 2nd, 3rd, 4th, &c., orders of differences by D1, D2, D3, D4, &c., and invert the order of the letters so that a shall stand first, we have

Here the coëfficients of a, b, c, d, &c., in the nth order of differences are evidently the coefficients of the terms of a binomial raised to the nth power; and their signs are alternately positive and negative; hence, when n is even, the first term of the nth order of differences is

a-nb+

_n(n−1). ̧__n(n−1)(n—2)d+, &c., and

1.2

1.2.3

—a+nb_n(n-1) c+n(n−1)(n−2)d—, &c., when n is odd.

1.2

1.2.3

Cor. It is evident from the coëfficients that when n=1, the value of D has only two terms, for then n-1=0; when n=2, this value has only three terms, for then n-2=0, and so on.

Д

Ex. 1. Find the first term of the fourth order of differences of the series 13, 23, 33, 43, 53, . . . . or 1, 8, 27, 64, &c.,

...

....

Here n=4, hence take five terms of the first value of D., and a=1, b=8, c=27, d=64, e=125, and D1=

4X3X2. 4×3×2×1×125=

X64+

1×2×3

1X2X3X4

1-32+162-256+125=0, Ans.

REMARK. It is evident the first term of any particular order of differences may be found by continued subtraction. It is important, however, that the learner should be acquainted with the general law as expressed in the above series.

EXAMPLES FOR PRACTICE.

...

...

2. Find the first term of the second order of differences of the series 12, 22, 32, 42, or 1, 4, 9, 16, 25. . . . Ans. 2. 3. What is the first term of the third order of differences of the series 1, 3, 6, 10, 15, &c.?

Ans. 0.

4. Required the first term of the fifth order of differences of the series 1, 3, 32, 33, 34, &c.

5. Find the first term of the fifth order of differences of the series 1,,, 76, &c.

Ans. —37.

ART. 326. PROBLEM II.— To find the nth term of the series a, b, c, d, e, &c.

From the preceding article we have seen that

It is evident from inspection that the coefficients of the first terms of the different orders of differences, in the value of any term of the series, as of e the fifth term, are the coëfficients of the terms of a binomial involved to a power whose exponent is one less than the number denoting the place of the term; that is, the coefficients of the nth term of the series, are the coefficients of the (n-1) power of a binomial. Hence, writing n-1 instead of n, in the coëfficients of the nth power of a+b, (Art. 319), the nth term of the series is

a+(n−1)D,+(n−1)(n—;

1.2

(n−1)(n—2)(n—3)D,+, &c.

1.2.3

Ex. 1. Find the 12th term of the series 1, 3, 6, 10, 15, 21, ..

1 3 6 10 15,

2 3 4 5,

1

0

1

and the succeeding orders of differences are also evidently 0; hence 12th term

=a+(n−1)D,+(n−1)(n—2)D,=1+11x2+ 11x10

1.2

=1+22+55=78. Ans.

2. Find the nth term of the series 2, 6, 12, 20, 30,

hence D1=4;

D2=2;

D3=0;

(n—1) (n—2)×2—n2+n. Ans.

1.2

From the formula n2+n, or n(n+1), any term of the series

is readily found; thus the 20th term =20(20+1)=420.

It is also evident that the nth term of a series can be found exactly only when some order of differences is zero.

3 Ex. a = 1, 2,1 = 3, D2 = 2, D2

8,

SERIES-DIFFERENTIAL METHOD.

48x. a = 1 20. 4 Dz

EXAMPLES

12th = 1 + 11×4+

3. Find the 15th term, and the nth term of the series 1, 22, 32,

42,.

. . or, 1, 4, 9, 16, .

Ans. 225, and n2.

4. Find the 12th term of the series 1, 5, 15, 35, 70, 126, &c.

1 + (n-1)2 + (+- Ans. 1365.
1+ moja y
of the series 1, 3, 6, 10, &c.

1.=2 mitt leve = 5. Find the nth term Dr=1,203=0

2

Ans. n(n+1)
n(n+1)

2

6. Find the nth term of the series 1, 4, 10, 20, 35, 56, &c.

Ans. n(n+1)(n+2)
2X3

7. Find the 9th term of the series 2·57, 4.7·9, 6·9 11, 8.11.13, &c.

Ans. 8694.

8. What is the nth term of the series 1×2, 3×4, 5×6, &c. ?
Ans. 4n2-2n.

ART. 327. PROBLEM III.- To find the sum of n terms of the series a, b, c, d, e, &c.

Assume the series 0, a, a+b, a+b+c, a+b+c+d,

Subtracting each term from the next succeeding, we have

a, b, c, d, e, &c.,

which is the series whose sum it is proposed to find. Hence, the
sum of n terms of the proposed series, which it is now required
to find, is the (n+1)th term of the assumed series.

It is evident the nth order of differences in the given series, is
equal to the (n+1)th order in the assumed series. Hence, if we
compare the quantities in the assumed series, with those of the
formula for finding the nth term of a series (Art. 326), we have

Substituting these values in the formula, we have 0+(n+1—1)a

(n+1—1)(n+1—2)

+

1.2

'D1+(n+1—1)(n+1—2)(n+1—3)D2+,

1.2.3

288

RAY'S ALGEBRA, PART SECOND. n' + 3n=0, me

or,

na+n(n−1) D1 + n (n−1)(n-2)D,+, &c., which is

1.2

1.2.3

the sum of n terms of the proposed series.

Ex. 1. Find the sum of n terms of the odd numbers 1, 3, 5,

Sum =na+n(n−1)D1=n×1+n(n−1) 2x2=n+n2—n=n2.

1.2

2

m+3

....

2. Find the sum of n terms of the series 12, 22, 32, 42, 52, . . . .
Here a=1, D,=3, D2=2, D ̧=0; hence,

3. Find the sum of n terms of the series 1+3+6+10+15,

&c.

2-3

4. Find the sum of 20 terms of the series 3+11+31+69 +131, &c.

Ans. 44330.

Ans. 53130.

6. Find the sum of n terms of the series of cube numbers 13+23+33+, &c. ̧ ̈ jik

7. Find the sum of n terms of the series 1+4+10+20

+35.

Ans. (n+1)(n+2)(n+3)
1X2×3×4

8. Find the sum of 25 terms of the series whose nth term is
n2(3n-2).

Ans. 305825.

ART. 328. PILING OF CANNON BALLS AND SHELLS.

Balls and shells are usually piled by horizontal courses, either in the form of a pyramid or a wedge; the base being either an equilateral triangle, or a square, or a rectangle. In the triangle and square, the pile terminates in a single ball, but in the rectangle it finishes in a ridge, or single row of balls.

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