2n

35. If a occurs in the expansion of (22+1),

, prove that its co

170. In the expansion of (1 + x)a the coefficients of terms equidistant from the beginning and end are equal.

The coefficient of the (r+ 1)th term from the beginning is "C,.

The (r+1)th term from the end has n + 1-(r+ 1), or n−r terms before it; therefore counting from the beginning it is the (n+1)th term, and its coefficient is "C which has been shewn to be equal to "C,. [Art. 145.] Hence the proposition follows.

171.

n-r

To find the greatest coefficient in the expansion of (1 + x)".

The coefficient of the general term of (1 + x)" is "C; and we have only to find for what value of r this is greatest.

;

By Art. 154, when n is even, the greatest coefficient is "C and when n is odd, it is "C, or "C +1; these two coefficients

being equal.

2-19

2

172. To find the greatest term in the expansion of (x + a)".

a

therefore, since x" multiplies every term in (1 + 2)",

it will be

sufficient to find the greatest term in this latter expansion.

Let the 7th and (r+1)th be any two consecutive terms. The (r+1)th term is obtained by multiplying the 7th term by n-r+ 1

a
;

a

that is, by ("+1-1).

[Art. 166.]

(r+1)th term is not always greater than the 7th term, but only

n+1

a

until -1) becomes equal to 1, or less than 1.

Now

х

so long as

1>

r

be an integer, denote it by p; then if r = p_the

multiplying factor becomes 1, and the (p + 1)th term is equal to the pth; and these are greater than any other term.

be not an integer, denote its integral part by q;

then the greatest value of r consistent with (1) is q; hence the (q+1)th term is the greatest.

Since we are only concerned with the numerically greatest term, the investigation will be the same for (x - a)"; therefore in any numerical example it is unnecessary to consider the sign of the second term of the binomial. Also it will be found best to work each example independently of the general formula.

find the greatest term in the expansion of (1+4x)8.

Denote the 7th and (r+1)th terms by T, and T+1 respectively; then

The greatest value of r consistent with this is 5; hence the greatest term is the sixth, and its value

Example 2. Find the greatest term in the expansion of (3- 2x)9 when

thus it will be sufficient to consider the expansion of (1-2).

Hence for all values of r up to 3, we have Tr+1>Tr; but if r=4, then Tr+1=Tr, and these are the greatest terms. Thus the 4th and 5th terms are numerically equal and greater than any other term, and their value

23

=36 × 84 × 8-489888.

H. H. A.

10

173. To find the sum of the coefficients in the expansion of (1 + x)".

In the identity (1+x)" =1+ C1x + С ̧x2 + С ̧3 + ... + С_x", put x=1; thus

COR.

2" =1+C,+C,+C+...+C

2

3

that is "the total number of combinations of n things" is 2" – 1. [Art. 153.]

174.

To prove that in the expansion of (1+x)", the sum of the coefficients of the odd terms is equal to the sum of the coefficients of the even terms.

H

In the identity (1+x)" 1+ C,x + С2x2 + C ̧3 + ... + С_x", - 1; thus

put x =

175. The Binomial Theorem may also be applied to expand expressions which contain more than two terms.

Example. Find the expansion of (x2+ 2x − 1)3.
Regarding 2x-1 as a single term, the expansion

= (x2)3 +3 (x2)2 (2x − 1) + 3x2 (2x − 1)2 + (2x − 1)3
=x+6x5+9x1 - 4x3 – 9x2+6x-1, on reduction.

+non)

The series (1) = (Co + C1 + C2+ · • +Cn) + (C1 + 2C2+3c3+

=2n+n

{1 + (n − 1) + (n − 1) (n − 2)

=2n+n (1+1)n—1

=2n+n. 2n−1.

1.2