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The preceding value is x-1, the resulting number of factors is 5, and their difference 1: hence by the rule,
Σux= (x−1) x (x+1) (x+2) (x+3) +c= (x-1)511+ c.
Ex. 2. Integrate (5x+1) (5x+6.)
Ex. 3. Given 3
Ex. 5. Integrate the expression (x+1)(x+3).
Ans. 32 (x−7)111 + c.
(4x-9) (4x-5) (4x-1)+c.
This expression not being composed of successive values, (the increment of x being generally taken in unity on account of the most frequent application of this method,) it must be reduced to such a form. We write, for this purpose, (x+1)(x+3)= (x+2) (x+3) — (x+3) each term of which is integrable by the and we have
Ex. 6. Integrate (2x+3) (2x+7); that is, (2x+5) (2x+7) −2 (2x+7).
Ex. 7. Prove (1) Σ (2x+1) (2x+3)2 = — (6x+7) (2x−1) (2x+1) (2x+3) + c.
(2x+1) (2x+3) (2x+5) (2x+3) (2x+5) (2x+1) (2x+3) (2x+5)
Ex. 10. Integrate
For in both cases Ax is constant. When Ax=1, we have simply,
VIII. To find the sum of a series, whose general term is given.
Write n+1 for n in the general term: then the integral is the sum of the series.
S1 = U+1; and
For let the series be u1 + u2 + U3 + +uS,: then u, + u2 + uz + Un + Un+1= S+1 Hence by subtraction AS, = integrating, S1 = ΣUn+1•
Ex. 1. Find the sum of the series 1 + 2 + 3 + +n.
By integration S1 = Σ (n+1)=
Hence AS, = n+1.
To find c, which is the same value whatever n may be, put n = 0; then S1 = also; and we have 0 0 + c, or c = 0. Whence the sum of the series of n terms is
n (n+1), as at p. 161.
Ex. 2. Find the sum of the series 12 + 2 + 32 +.... +n2.
Here un2 and unt1 = (n+1)2. Whence as before
= (n+1)2, and integrating, S1 = Σ (n+1)2 =
the correction being found O as in the last example.
Ex. 3. Sum the series of cubes 13 + 23 + 33 + ....n3.
n (n+1) (2n+1)
`n (n- +1
Here 4S„ = (n+1)3, and S„* =
Ex. 4. Find the sum of n terms of the series 1.2 + 2.5 + 3.8 + 4.11 +........ Here the general term of the first factor of the several terms is obviously n, and the second (found by III.) is 3n-1. Hence the general term of the series
* By comparing the solutions of examples 1 and 3 we see that the sum of n terms of the cubes of the natural numbers is equal to the square of the sum of those numbers themselves.
is un = n (3n—1); and the increment or (n+1)th term is u2+= (n+1) (3n+2) =3n (n+1) + 2(n+1). Whence integrating, S„ = n2 (n+1).
Ex. 5. Show that 12 +32 + 52 + 72 +.. + (2n−1)2 = n (4n2-1).
Ex. 6. Find the sum of n terms of each of the following series:
Here S1 = Σ
Ex. 7. Find the sum of the 2nd, 3rd, and 4th powers of the preceding series of numbers.
Ex. 8. Sum the series of n terms of
Here un =
(2n+1) (2n+2) (2n+3) (2n+4)*
(2n-1) 2n(2n+1) (2n+2) Now the denominator of the function u2+, not being composed of consecutive integer values of n, it must be so transformed as to answer that condition, or else into some other form which can be integrated. The latter plan is adopted here.
Put 2n+1=u, then 2n + 3 = u1 and 2n+2v, then 2n + 4 = v1 where u, and v1 are the next values of u and v. uAv + vAu + Au Av uuvv
Au and Av in (1) and insert them in this equation, we get
Hence integrating, restoring the values of u, v, u1, v1, and correcting we finally obtain the sum of the given series, viz. :
Ex. 5. Find the sum of fifteen terms of each of the following series :
IX. The summation of series whose general term is not known.
The sum is obtained from the following expression, by the insertion of the values of n and the first terms of the several orders of differences, found as in
* The several series in this example have, in connection with each other, very remarkable properties, and have been much used in mathematical research. They are called the figurate series, or series of figurate numbers; and are thus derived from each other by successive additions. The first row is a series of units, and the others are formed in succession by adding each term of the mth row to the (m+1)th term of the (m—1)th row. The summation shows that S in the successive series is expressed by the successive terms of the expanded binomial (1+1)" or (1-1)-"; though only that of the mth series is put down. See also P. 289.
Whence, adding the several vertical columns, (which are the same with the series in Ex. 11, VIII. p. 285,) we have
Ex. 1. To find the sum of n terms of 1+ 2+ 3+ 4
second and all higher differences.
Hence u1 = 1, Au1 =
1, Au1 = 0, and so on: and we have
n(n - 1) n(n + 1)
as was obtained by integration at p. 283.
Ex. 2. Find the sum of 12 + 22 + 32 + +102.
Hence u1 = 1, Au1 = 3, A2u1 = 2, ▲3u, = 0, and so on; and
Ex. 3. Find the sums of 50 terms of the series in IV. Examples 2, 3, 6. It has been seen (p. 275) that an expression of the nth degree has its nth differences equal, and all the higher orders in succession become 0: but when an expression is of any other form, (as a* or log. ≈ for instance,) the successive differences never vanish; though in many of them, these differences become very small, and the assumption of their becoming 0, leads to a very small amount of error in the final numerical result. When the sum of n terms of such a series, therefore, whose differences in successive orders become very small, is required, we are at liberty to assume those differences as actually vanishing.
Ex. 4. Required the sum of all the logarithms on the fifteen pages of Hutton's Tables, from p. 186 to p. 201, inclusive, the entire numbers being integer. Here n=3000, u1=5, Au1='00000435, A3u,='00000000005 *. Hence S8000 = 8000.5 +
00000000005 =40143-4476668; which is, probably, true to five decimal places.
Let Иде un, be the given values of an unknown function u when equidistant values of x are substituted, and any number, p, of them be absent: it is required to supply the absent terms, and to find the value of the function u, for any value of x intermediate between its extreme given values.
The second difference is taken at
of the interval 00000001, this being the dif
ference at the 220th, 210, 200th, 190th, and 180th, terms: or nearly a mean of all.