Represent, for shortness, any series of the form

Now, if the series on the right of (i) and (ii) be multiplied, and the product be arranged according to ascending powers of x, the result must involve m and n in the same way whatever their values may be. But, when m and n are positive integers, we know that ƒ (m) is (1+x)”, and that ƒ (n) is (1+x)”, and the product ƒ (m) × ƒ (n) is therefore (1+x)+n, which again, as m + n is a positive integer, is f(m+n). Hence when m and n are positive integers the product f (m) × ƒ (n) is ƒ (m + n); and, as the form of the product is the same for all values of m and n it follows that

f(m) × ƒ (n) = f (m + n)............(a),

for all values of m and n. [See however Art. 276.]

From this point the proof is the same as in Art. 279.

Ex. 1. Expand (1+x)−1.

Put n=1 in the above formula; then we have

(-1) (-2) (-1) (-2) (-3)
x2 +

1.2

( − 1) ( − 2).............. ( − r)

+

r

=1−x + x2 - x3+

1.2.3

· + ( − 1)" x" + .....

This example illustrates the necessity of some limitation in the value of x; for we know [Art. 225] that 1−x+x2- .

is not

equal to

1
1+x

Here again it is clear that the result cannot be true for all values of x; if x=2, for example, we should have

All the terms are positive, for in the general term there are 2r negative factors.

xr+

2.4.6

2r

5

Ex. 5. Expand (a3 – 3a2x) according to ascending powers of x.

After the second, all the signs are positive; for in the general term there are r−2+r, that is an even number, of negative factors.

281. The (r+1)th term of the expansion of (1 + x)"

is obtained from the rth by multiplying by

n

r+1

X,

negative if n + 1 is negative; and, whatever n + 1 may be,

-1+

n+1

r

will be negative for all terms after the first for which r> n + 1.

Hence, if x be positive, the ratio of the r + 1th and rth terms will be always negative when r>n+1. The terms of the expansion of (1+x)" will therefore be alternately positive and negative after r terms, where r is the first positive integer greater than n + 1.

If x be negative, the ratio of the (r+1)th and rth terms will be always positive when r>n+1. The terms of the expansion of (1-x)" will therefore be all of the same sign as the rth term, where r is the first positive integer greater than n+1; and, as a particular case, all the terms of the expansion of (1-x)" are positive when n is negative.

For example, all the terms in the expansion of (1 - x) are of the same sign as the rth, where r is the integer next greater than +1, so that r is 3. Also, after the ninth, the terms of the expansion of (1+x)131⁄23 are alternately positive and negative.

15

282. Greatest Term. In the expansion of (1 ± x)" by the binomial theorem, we know that the ratio of the

(r+1)th term to the rth is +

Fx1

n+1

r

n―r + 1

r

X, that is

; we also know that x must be numeri

cally less than 1, unless n is a positive integer.

First suppose that n+1 is negative, and equal to -m. Then the absolute value of the ratio of the (r+1)th

term to the rth term is a (1+m).

is

ANY

Hence the rth term

m

(r+1)th term according as x (1+1; that is,

according as r

Hence, if

r

(1 + n) x

that is "

(1 + n) x

1. 30

< 1.

Ꮳ

be an integer, r suppose, the rth

term will be equal to the (r+1)th term, and these will

- (1 + n) x

be greater than any other terms. But, if 1 ac be not an integer, the rth term will be the greatest when (1 + n) a

Next, suppose that n + 1 is positive, and let k be the integer next greater than n + 1. Then, if r be equal or n+1

greater than k, -1 will be negative and less than

unity; hence, as a must be less than unity, each term after the kth will be less than the one before it, and therefore the greatest term must precede the kth. And since, for values of r less than n+1,

positive; the rth term will be (+1)th according as

(n + 1 − 1) a = 1; that is, according as r

r

Hence, if

>

(n + 1) x

1 + x

> (n + 1) x

< 1+x

be an integer, r suppose, the rth

term will be equal to the (r+1)th, and these will be

other terms. But, if

greater than any other terms.

(n + 1) x

be not an

1 + x

integer, the rth term will be the greatest when r is the

Ex. 1. Find the greatest term in the expansion of (1 − x)−1, when

= 4. Hence the

one another, and are greater

Ex. 2. Find when the expansion of (1-x) begins to converge, if

3

convergence begins after the 23rd term.

Ex. 3. Find the greatest term in the expansion of (a + x)11⁄2, when

1. Find the general term in the expansion of each of the following expressions by the binomial theorem.

(i) (1-x)-3, (ii) (1-x)-3,

(iii) (1 − x) ̄”.