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•: (5c2 — 2yz)* = 625 c3 · 1000c yz + 600 c1 y2 z2 — 160 c2 y3 z3 + 16 yʻz1

111. We have sometimes occasion to employ a particular term in the expansion of a binomial, while the remainder of the series does not enter into our calculations. Our labour will, in a case like this, be much abridged, if we can at once determine the term sought, without reference either to those which precede, or to those which follow it. This object will be attained, by finding what is called the general term of the series.

If we examine the general formula, we shall soon perceive that a certain relation subsists between the coefficients and exponents of each term in the expanded binomial, and the place of the term in the series; thus,

The first term is " which may be put under the form −1+1

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Observing the connection between the numerical quantities, it is manifest, that if we designate the place of any term by the general symbol p, the pth term is,

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This is called the general term, because by giving to p the values 1, 2, 3, 4, we can obtain in succession the different terms of the series for

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Substituting these values in the general expression, we find that the term

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all the even

Since the second term of the proposed binomial has the sign terms of the expansion will have the sign therefore the 5th term is,

and all the odd terms the sign+;

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Example IX.

Required the middle term of the expansion of (r~~α) 18.

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Since the exponent is 18, the whole number of terms will be 19, and hence the middle term will be the 10th; and since it is an even term, it will have the sign; hence it will be,

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112. By employing the Binomial Theorem, we can raise any polynomial to any power, without the process of actual multiplication.

For example, let it be required to raise x+a+b to the power of 5.

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= x+4x3y+6x2 y2+xy3+y, putting for y its value,
x2 + 4x3 (u+b) + 6x2 (a+b)3 + 4x(a+b)3 + (a+b)*

Expanding (a+b)2, (a+b)3, (a+b), by the Binomial Theorem, and performing the multiplications indicated, we shall arrive at the expansion of (x+a+b)*.

It is manifest, that we may apply a similar process to any polynomial.

113. In the observations made upon the expansion of (x+a) ", we have supposed n to be a positive integer. The binomial theorem, however, is applicable, whatever may be the nature of the quantity n, whether it be positive or nega tive, integral or fractional. ▾ When n is a positive integer, the series consists of n+1 terms; in every other case the series never terminates, and the developement of (x + a)" constitutes what is called an infinite series.

Before proceeding to consider this extension of the theorem, we may remark that in all our reasonings with regard to a quantity, such as (x + a)", we may

• No algebraist has succeeded in proving this in a manner altogether satisfactory. The least ex ceptionable of the demonstrations which have been proposed, will be given in a subsequent chapter.

corfine our attention to the more simple form (1 + a)" to which the former may always be reduced. For,

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Suppose n=— where r and s are any whole numbers whatever,

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Then (z + a) * becomes (x + a), and substituting for n in the series.

(x + a); = x¦ (1 + 2 );

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Or reduced,

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r (r—s) (r — 2s)

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r (r − s) (r — 2 s) (r — 3 s) a*

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114. The binomial theorem, under this form, is extensively employed in analysis for developing algebraic expressions in series.

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=

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•− 2 (p + s ) ( ' + ' ';) ~ '. Here, r = −3,1 = 4

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