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and it will give the root required: observing that p denotes the first term, e the second term divided by the first, ñ the index of the power or root; and A, B, C, D, &c, denote the several foregoing terms with their proper signs.
1. To extract the sq. root of a +b?, in an infinite series.
1 Here p = a , e
(a?)* = (a?)* =a= A, the 1st term of the series.
62 - AQ = I xa x
= B, the 2d term.
= C, the 3d term 2n
2a al 2.4a3 - 2n
= D the 4th. 3n
3.86 Hence at
&c, or 2a 2.403 2.4.6as 62 64 66 568 ht +
&c, is the series required. 2a 8a3
2. To find the value of or its equal (a − x), in
(a - x)? an infinite series *.
* Note. To facilitate the application ‘of the rule to fractional examples, it is proper to observe, that any surd may be taken from the denominator of a fraction and placed in the numerator, and vice versa, by only changing the sign of its index.
1 =1 X x2 or only rr?; and =1X (a + b) or
ci (a + b)-; and =u' (a + x); and =rixi'}; also
(a + x)* (cle + x') } (1931 = (a +x®)+ x (a® — 227?; &c.
-2 Heré P =a, q=
-2; theref. 1 pö = (a)-- = a^2 = = A, the 1st term of the series.
2x -AQ=-2 X
= 2a-3x = B, the 2d term. n
2.x X 3.72 BQ = - }*
= 3a-4x2 = c, the 3d. 2n
at m - 2n
= 48-5.73 = D. 3n
as Hence ? + 20-3x + 3a 4x + 40-5x3 + &c, or 1 2x 3 3х2 4.23 5.24 + + +
&c, is the series required. a4 as
1 4. To expand v
in a series. (a + c) (a2 + x2)
1 x2 Ans. +
&c. ? 2a3 sas
a 5. To expand
in an infinite series. (a - b)2
26 362 463 564 Ans. 1+ + + + &c.
G. To expand Vã -- .zor (a? — xo); in a series.
7. Find the value of 3/(a3 – 63) or (a} -- 63); in a series.
$. To find the value of s/(as +x5) or (as +xs) in a series.
&c. 5a4 2549 125a14
a? 62 9. To find the square root of in an infinite series.
a? + b2
b x? 23
&c. a 2a?
u + b3
ARITHMETICAL PROPORTION is the relation between two numbers with respect to their difference.
Four quantities are in Arithmetical Proportion, when the difference between the first and second is equal to the difference between the third and fourth. Thus, 4, 6, 7, 9, and a, a + d, b, b + d, are in arithmetical proportion.
Arithmetical Progression is when a series of quantities have all the same common difference, or when they either increase or decrease by the same common difference. Thus, 2, 4, 6, 8, 10, 12, &c, are in arithmetical progression, hav-, ing the common difference 2; and a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, &c, are series in arithmetical progression, the eommon difference being d.
The most useful part of arithmetical proportion is contained in the following theorems :
1. When four quantities are in Arithmetical Proportion, the sum of the two extremes is equal to the sum of the two
Thus, in the arithmeticals 4, 6, 7, 9, the sum 4 + 9 = 6+7= 13: and in the arithmeticals a, a +d, b, b+d, the sum a +b+d=a + b + #.
2. In any continued arithmetical progression, the sum of the two extremes is equal to the sum of any two terms at an equal distance from them.
1, 2, 3, 4, 5, 6, &,
Thus, if the series be 1, 3, 5, 7, 9, 11, &c.
Then 1+11= 3 + 9 = 5 + 7 = 12. 3. The last term of any increasing arithmetical series, is equal to the first 'term increased by the product of the common difference multiplied by the number of term's less one; but in a decreasing series, the last term is equal to the first term lessened by the said product.
Thus, the 20th term of the series 1, 3, 5, 7, 9, &c, is = 1+2 (20-1) = 1 + 2 19 =1+ 38 =*39.
And the nth term of a, a-d, a-2d, a-3d, a-4d, &c, is = a + (n-1) d = a-(n-1)d.
4. The sum of all the terms in any series in arithmetical progression, is equal to half the sum of the two extremes multiplied by the number of terms. Thus, the sum of 1, 3, 5, 7, 9, &c, continued to the 10th (1 + 19) x 10 20 x 10
I 10 X 10 = 100. 2
2 And the sum of n terms of a, atd, a + 2d, a +3d, to a + md, is = (a + a + md). = (a + 1md) n,
mdikis filminino's EXAMPLES FOR PRACTICE. '
Siitoi 1. The first term of an increasing arithmetical series is 1, the common difference 2, and the number of terms 21; required the sụm of the series?? First, 1 + 2 X 20 1 + 40 = 41, is the last term.
1+ 41 Then * 20 = 21 20 = 420, the sum required.
2 2. The first term of a decreasing arithmetical series is 199, the common differences, and the number of terms 37 ; required the sum of the series ? ; First, 199-3.66 = 199 - 198 =), is the last term.
199 + 1 Then
X 67 = 100 X 67 = 6700, the sum re
lui quired. 3. To find the sum of 100 terms of the natural numbers
Ans, 5050. VOL. I.
4. *Required the sum of 99 terms of the odd numbers 1, 3, 5, 7, 9, &c.
· Ans. 9811. 5. The first term of a decreasing arithmetical series is 10, the common difference }, and the number of terms 21; required the sum of the series?
Ans. 140. 6. One hundred stones being placed on the ground, in a straight line, at the distance of 2 yards from each other; how far will a person travel, who shall bring them one by one to a basket, which is placed 2 yards from the first stone?
Ans. 11 miles and 840 yards.
A TRIANGULAR Battaliont, consisting of thirty ranks, in which the first rank is formed of one man bnly, the
* The sum of any number (n) of terms of the arithmetical series of odd number 1, 3, 5, 7, 9, &c, is equal to the square (no) of that number. That is, If b, 3, 5, 7, 9, &c, be the numbers, then will
12; 2, 3, 4, 5%, &c, be the sums of 172, 3, &c, terms.
1. 3 = 4 or 2, the sum of 2 terms,
9 + 7 = 16 or 4, the sum of 4 terms, &c. For, by the 3d theorem, 1 + 2 (n-1) = 1 + 2n-2 = 2n-1 is the last term, when the number of terms is n; to this last term 2n-1, add the first term 1, gives 2n the sum of the extremes, or n half the sum of the extremes ; then, by the 4th theorem, nxn
no is the sum of all, the terms. Hence it appears in general, that half the sum of the extremes, is always the same as the number of the terms n; and that the sum of all the terms, is the same square of the same number, n2.
on Arithmetical Proportion in the Arithmetic,
t By triangular battalion, is to be understood, a body of troops ranged in the form of a triangle, in which the ranks exceed each