38. Prove that the nth coefficient of the expansion of (1-x) ̄”

is always the double of the (n - 1)th.

39. Shew that if t denote the middle term of (1 + x)3,

t + t + t ̧ + ...... = (1 − 4x) ̄§.

1

then

41. Find the sum of the squares of the coefficients of the expansion of (1+x)", where n is a positive integer.

P2n+1+P1P2n+PaPan-1+......+P2-1P+2+PP2+1=

43. Prove that the coefficient of x in the expansion of

is equal to the coefficient of x" in the expansion of

44. Find the coefficient of x in

(1 + 2x + 3x2 + 4x3 + ad inf. ......)".

XXXVII. THE MULTINOMIAL THEOREM.

528. We have in the preceding chapter given some examples of the expansion of a multinomial; we now proceed to consider this point more fully. We propose to find an expression for the general term in the expansion of (a + a ̧x+α ̧x2 + ɑzÃ3 + ......)". The number of terms in the series a, a,, α, may be whatever, and n may be positive or negative, integral or fractional.

......

any

Put b1 for ax+а ̧x2 + α2x3 + then we have to expand (a+b)"; the general term of the expansion is

μ being a positive integer. Put b, for ax2+a,x3+...... then b,μ= (ax+b)"; since μ is a positive integer the general term of the expansion of (ax + b)" may be denoted either by

we will adopt the latter form as more convenient for our purpose.

Combining this with the former result, we see that the general term of the proposed expansion may be written

and the general term of the expansion of this will be

Hence the general term of the proposed expansion may be written

Proceeding in this way we shall obtain for the required general term

If we suppose n-μ=p, we may write the general term in

Thus the expansion of the proposed multinomial consists of a series of terms of which that just given may be taken as the general type.

It should be observed that q, r, s, t, are always positive integers, but p is not a positive integer unless n be a positive integer. When p is a positive integer, we may, by multiplying both numerator and denominator by [p, write the coefficient n (n-1) (n-2)...... (p + 1)

grst......

in the more symmetrical form

|n

p q r s t

$529. Suppose we require the coefficient of an assigned power of x in the expansion of (a + a ̧x + a ̧x2 + ..............)", for example, that of ". We have then

We must find by trial all the positive integral values of q, r, s, t, which satisfy the first of these equations; then from the second equation p can be found. The required coefficient is then the sum of the corresponding values of the expression

n(n-1) (n-2)...... (p+1)
Lars Lt......

When n is a positive integer then p must be so too, and we may use the more symmetrical form

530. For example, find the coefficient of x in the expansion

of (1 + 2x + 3x2 + 4x3)*.

Begin with the greatest admissible value of s; this is 8 = 2, with which we have r = 0, q = 1, p = 1. Next try s=1; with this we may have

0

2

0 1 3

0

Thus the required coeffi

r = 2, q = 0, p= 1; also we may have r=1, 10
q=2, p=0. Next try 8=0; with this we may
have r = 3, q= 1, p=0. These are all the so-
lutions; they are collected in the annexed table.
Also a = 1, a1 = 2,

cient is

that is,

14

14

=

= 4.

384 + 432 + 576 +216; that is, 1608.

Again; find the coefficient of a3 in the expansion of

All the solutions are given in the annexed table, and the required coefficient is

the proposed expression is {(1-x)-2, that is, (1-x)-1. And

(1 − x) ̄1 = 1 + x + x2 + ∞3 + ... ... ;

thus we see that the coefficient of x3 ought to be 1; and the student may exercise himself by applying the multinomial theorem to find the coefficients of other powers of x, as, for example, x*.

EXAMPLES OF THE MULTINOMIAL THEOREM.

Find the coefficients of the specified powers of x in the following expansions: