180. To prove negative quantity.

the Binomial Theorem when the index is any

It has been proved that

f (m) xf(n) = f (m+n)

for all values of m and n. Replacing m by - N positive), we have

ƒ(-n) ׃(n) = f (− n + n)

(where n is

=ƒ (0)

= 1,

since all terms of the series except the first vanish;

but f(n)=(1+x)", for any positive value of n;

which proves the Binomial Theorem for any negative index. Hence the theorem is completely established.

181. The proof contained in the two preceding articles may not appear wholly satisfactory, and will probably present some difficulties to the student. There is only one point to which we shall now refer.

In the expression for f(m) the number of terms is finite when m is a positive integer, and unlimited in all other cases. See Art. 182. It is therefore necessary to enquire in what sense we

are to regard the statement that ƒ (m) × ƒ (n) = f(m + n). It will be seen in Chapter XXI., that when a< 1, each of the series f(m), f(n), ƒ (m + n) is convergent, and f(m + n) is the true arithmetical equivalent of f(m) ×f(n). But when x>1, all these series are divergent, and we can only assert that if we multiply the series denoted by f(m) by the series denoted by ƒ (n), the first r terms of the product will agree with the first r terms of ƒ (m + n), whatever finite value r may have. [See Art. 308.]

182. formula

In finding the general term we must now use the

written in full; for the symbol "C, can no longer be employed when n is fractional or negative.

Also the coefficient of the general term can never vanish unless one of the factors of its numerator is zero; the series will therefore stop at the 7th term, when n−r+1 is zero; that is, when r=n+1; but since r is a positive integer this equality can never hold except when the index n is positive and integral. Thus the expansion by the Binomial Theorem extends to n + 1 terms when n is a positive integer, and to an infinite number of terms in all other cases.

Example 1. Find the general term in the expansion of (1+x)2,

The number of factors in the numerator is r, and r- 1 of these are negative; therefore, by taking - 1 out of each of these negative factors, we may write the above expression

Example 2. Find the general term in the expansion of (1 – nx).

Example 3. Find the general term in the expansion of (1-x)-3.

by removing like factors from the numerator and denominator.

183. If we expand (1-x)-2 by the Binomial Theorem, we obtain

(1 − x)2 = 1 + 2x + 3x2 + 4x3 + ;

but, by referring to Art. 60, we see that this result is only true when x is less than 1. This leads us to enquire whether we are always justified in assuming the truth of the statement

and, if not, under what conditions the expansion of (1 + x)" may be used as its true equivalent.

in this equation put x = 2; we then obtain

take

This contradictory result is sufficient to shew that we cannot

as the true arithmetical equivalent of (1 + x)" in all cases.

Now from the formula for the sum of a geometrical progression, we know that the sum of the first r terms of the

series (1)

1 - x

and, when a is numerically less than 1, by taking

large we can make

1 Ꮳ

sufficiently

as small as we please; that is, by taking

a sufficient number of terms the sum can be made to differ as

It will be seen in the chapter on Convergency and Divergency of Series that the expansion by the Binomial Theorem of (1+x)" in ascending powers of x is always arithmetically intelligible when x is less than 1.

But if x is greater than 1, then since the general term of the series

1 + nx +

n (n-1)