BINOMIAL THEOREM.

We have already found by actual multiplication (Art. 96),

that

(a+x)1=a+x

(a+x)2=a2+2ax+x2

(a+x)3=a3+3a2x+3ax2+x3

3

(a+x)=a*+4a3x+6a2x2+4ax3+x1

(A)

Now, the BINOMIAL THEOREM teaches us the law by which we may write the development of (a+x)" for any values of a, x, and m.

To determine this law, assume

m

(a+x)ñ=A+A1x+A2x2+A3x3+&c.

(1)

We have taken the exponent of this binomial fractional in order to make the development more general.

We know by the principle of (Art. 182), that this assumed value (1) must be true for all values of x; when x=0

m

=

it becomes an A, introducing this value of A in (1), it be

(a+x)ñ=añ+A ̧x+A2x2+A3x3+&c.

In (2), writing x1 for x, and it becomes

(2)

(a+x1)ñ=añ+A1≈1+A2x}+A ̧x}+&c. (3)

Subtracting (3) from (2), we find

}

Dividing the left-hand member of (4) by u”—u1", and the right-hand member by its equal x-x1, observing to sub

stitute uTM—u ̧TM for (a+x) −(a+x1)", as given by (6), and it will become

Dividing both numerator and denominator of the left-hand

membe of (8) by u―ú1, and performing the divisions indicated in the right-hand member, and we obtain by the aid of equation (B), Art. 184, the following:

2

A1+A2(x+x1)+A3(x2+xx1+x})+&c.

2

Now, in (9), suppose x=x1, and consequently u=u1, and

=A1+2A2x+3A ̧x2+441x3+&c. (10)

Resubstituting (a+x)" for u in (10), and it will become

m

n

m

(a+x)"_A ̧+2A ̧x+3A ̧x2+44 ̧1⁄23+&c.

a+x

1

Multiplying through by a+x, and we obtain

(11)

Equating the right-hand members of (13) and (12), we have

Now, by the principle of (Art 182), we must equate the coefficients of like powers of x, by which means we have

These values of A1; A2; Aз; &c., substituted in (2),

The coefficient of the (p+1)th term as given by (15), be

m

n

(185.) The numerator of this coefficient being formed of factors decreasing regularly by one, it follows that when p=m+1 it will' vanish, and then the series must terminate; so that the number of terms of the expansion (B) will be m+1. But when is fractional, or a negative integer, the number of terms of the expansion must be infinite.

When a os x becomes negative, then those terms of the expansion will change signs, which contain odd powers of this negative quantity. (186.) If in (B), we write a for x and x for a, we shall have

m(m-1) (x+a)"="+mxm-la+

m-2a2 + &c.

(17)

a

m

1

2

Now, since the left-hand members of (B) and (17) are evidently equal, their right-hand members must be; and since, when m is a positive integer the number of terms of (B) as well as (17) is equal to m+1, it follows that the terms of the expansion (B) must be homogeneous and symmetrical, and therefore of this form

If in (A) we suppose a=x=1, we shall find

Therefore, in any expansion of a binomial, whose terms are both positive, the sum of the coefficients is equal to the same power, or root of 2.