211. Problem. To find any power of a binomial.

Solution. This power might be obtained directly by multiplication, but the operation is long and tedious, and can be avoided by a process invented by Newton. To obtain this process, let the given binomial be a+x, let n be the exponent of the power, and let the product be arranged according to the powers of x, so that

(a + x)” = A + B x + C x2+D x3 + E x2 + &c., in which the coefficients, A, B, C, &c., are to be determined; none but positive integral powers of x are written in the second member, because the product could, evidently, give no others, and all the positive integral powers of x are included, because the coefficients of any which are superfluous must be found to vanish.

First. To find the value of A. Let

x=0

and the development becomes

an A.

Binomial Theorem.

Secondly. To find the form in which a enters into the development. Let

and let the corresponding values of a", B, C, &c., be 1, B', C', &c., and we have

(1+x')=1+ B' x' + C' x' 2 + D'x'3 + &c.

in which A', B', C', &c., must be independent of a. The product of this equation by an is

a" (1+x')"=(a+a x')"=a^+B' a2 x' +C′ aˆ x2+&c., in which, if we put

we have

and

ax' = x, or x' =

x'

a^ x' = an−1x, a" x12 = an−2 x2, &c.,

(a+x)=a+B' a"-1 x + C' a"−2x2+D'aa¬3 x3+&c. Thirdly. To find the coefficients. The derivative of the last equation is, by examples 4 and 8 of art. 175, n(a+x)-1=B'a"-1+2C"a"-2x+3D'a-3x2+4E'an-473

&c.,

which, multiplied by (a+x), gives

n(a+x)"=B'a"+2C'a2¬1 x+3D'a2¬2x2+4E' a2¬3x3+&c. +B'an 1x+2C' a"-2x2+3D' a"-3x3+&c.

The product of (a + x)" by n gives, also,

n(a+x)" =na"+n B'a"-1x+nC"a"-2x2+n D'a-3 x3+&c. which, compared with the preceding equation, gives, by art. 163,

B'n,

2C+B'=n B', or 2 C'=(n-1)B', or C'={(n−1)B'; 3D'+2C=nC', 3D'=(n-2) C',

4E+3 D=nD', 4E' (n-3)D',

D'={(n—2)C′;

E'=1(n-3)D';

&c.

Binomial Theorem.

The combination of these results gives

Sir Isaac Newton's Binomial Theorem.

The first term of any power of a binomial is the same power of the first term of the binomial.

In the following terms of the power the exponent of the first term continually decreases by unity, whereas the exponent of the second term of the binomial, which is unity in the second term of the power, continually increases by unity.

The coefficient of the second term of the power is the exponent of the power.

If the coefficient of any term is multiplied by the exponent which the first term of the binomial has in that term, and divided by the place of the term, the result is the coefficient of the next following term.

212. Corollary. The equations of the preceding article give

Binomial Theorem.

213. Corollary. If x is changed into

ceding formula, it becomes

the signs of every other term being reversed.

214. Corollary. The preceding formula, written in the reverse order of its terms, gives

-1

n (n − 1) q2 xn−2+ &c. (x+a)" =x+naxn−1+ 2

whence it appears that

The coefficients of two terms which are equally distant, the one from the first term, and the other from the last term, are equal.

But, by the above formula,

(x − y)6 — x6 — 6 x5 y + 15 x1 y2 — 20 x3 y3 +15 x2y1

in which, if we substitute the values of x and y, we have

Binomial Theorem.

(2 ab-2c-4bc2 d)6 = 64 a6 b-12 c6-48 a5 b-9 c7 d+ 15 a4 b-6 c8 d2a3b-3 c9 d3 + 15 a2 c10 d4

3

a b3 cll d5 + 4096 b6 c12 dε.

2. Find the 10th power of a+b.

Ans. a10+10 a9 b+45 a8 b2+120 a7 b3+210 a6 b1+ 252 a5 b5 +210 a4 b6+120 a3 b7+45 a2 b® +10 a b9 + 610

3. Find the 11th power of 1 — x.

Ans. 1-11x+55 x2-165 x3+330 x1—462x5462 x6 -330x7165x8-55 x9 +11 x10x11.

4. Find the 4th power of 5-4 x.

Ans. 625-2000x2400 x2-1280 x3+256 x1.

5. Find the 7th power of x+2y.

Ans. 17+31⁄2 x6 y + 21 x5 y2+ 35 x1 y3 + 70 x3 y1

+168 x2 y5+224 x yo + 128 y7.

6. Find the 4th power of 5 a2 c2 d· 4 ab d2.

Ans. 625 a8 c8 d4 — 2000 a b c d5 + 2400 ao b2 c4 d¤

– 1280 a5 b3 c2 d2 +256 a1 ba d3.

216. Problem. nomial.

To find any power of a poly

Solution. Suppose the terms of the given polynomial to be arranged according to the powers of any letter, as x, as follows;

a + bx + c x2+dx3 + ex2 + &c.,

in which the successive coefficients are denoted by the successive letters of the alphabet. The following is

Arbogast's rule for finding any power of the poly

nomial.