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381. If in equation (1) of the preceding article we suppose b=c=d=a, we obtain

(x+α)1= x2+4ax3+6a2x2 + 4a3x+aa.

We shall now employ the same method to prove a formula known as the Binomial Theorem, by which any binomial of the form x+a can be raised to any assigned positive integral power. 382. To find the expansion of (x+a)" when n is a positive integer.

Consider the expression

(x+a)(x+b)(x+c)......(x+k),

the number of factors being n.

The expansion of this expression is the continued product of the n factors, x+α, x+b, x+c,......x+k, and every term in the expansion is of n dimensions, being a product formed by multiplying together n letters, one taken from each of these n factors.

The highest power of x is ", and is formed by taking the letter x from each of the n factors.

The terms involving -1 are formed by taking the letter from any n-1 of the factors, and one of the letters a, b, c,...k from the remaining factor; thus the coefficient of x-1 in the final product is the sum of the letters a, b, c,......k; denote it by S1

The terms involving -2 are formed by taking the letter from any n-2 of the factors, and two of the letters a, b, c,...k from the two remaining factors; thus the coefficient of x-2 in the final product is the sum of the products of the letters a, b, c,...k taken two at a time; denote it by S.

And, generally, the terms involving an-rare formed by taking the letter from any n-r of the factors, and r of the letters a, b, c,...k from the r remaining factors; thus the coefficient of am-r in the final product is the sum of the products of the letters a, b, c,...k taken r at a time; denote it by S.

The last term in the product is abc...k; denote it by S

Hence

(x+a)(x+b)(x+c)..............(x+k)

=x+S1x2-1+Sqx2-2+...+S,xn+...+Sn-1x+Sn

In S, the number of terms is n; in S, the number of terms is the same as the number of combinations of n things 2 at a time; that is, "C2; in S, the number of terms is "C3; and so on.

Now suppose b, c,...k, each equal to a; then S, becomes "C1a; S2 becomes "Ca2; S becomes "Ca3; and so on; thus (x+α)=x+C2axn−1+1С2α2x2-2+nCza3xn−3+...+"C'nan;

substituting for "C1, "C2,...we obtain
n (n − 1) a2xn−2
(x+a)n = x2+nax2-1+'

1.2

+

n(n-1) (n-2)
1.2.3

the series containing n+1 terms.

This is the Binomial Theorem, and the expression on the right is said to be the expansion of (x+a)".

If we write a in the place of a, we obtain (x-a)n = x2+C1 (-a),xn−1 +С2 (-a)2xn-2

383. The coefficients in the expansion of (x+a)" are very conveniently expressed by the symbols "C1, "C2, "C3,..."Cn. We shall, however, sometimes further abbreviate them by omitting n, and writing C1, C2, C3,... Cn. With this notation we have (x+α)=x+C1ax-1+C2α2x2+С2α3μ3¬3+...+Сnan.

-2

a3n-з+...+an,

+C3(-a)3x-3+ ... + C2 (-a)TM = x2 - С1axn−1+С2а2x2¬2 — С2а3ñ3¬3+...+(−1)nСnan.

Thus the terms in the expansion of (x+a)" and (x − a)" are numerically the same, but in (-a)" they are alternately positive and negative, and the last term is positive or negative according as n is even or odd.

Example 1. Find the expansion of (x+y).
By the formula, the expansion
=x+°C1x3y+®Cqx1y2+°Сqx3y3+°C+x2y*+®C5x¥3+°C ̧¥
=x+6x3y+15x4y2+20x3y3 + 15x2y1+6xy5+yo,

on calculating the values of "C1, 6С2, 6С3,.......

Example 2. Find the expansion of (a – 2x)7. (a -2x)=a7-7С1ao (2x)+7С2a3 (2x)2 − 7Сça1(2x)3 +

to 8 terms.

Now remembering that "C,="C-, after calculating the coefficients up to 7C, the rest may be written down at once; for 7C4=7C3; "C=C2; and so on. Hence

7.6

(a -2x)7=a7 - 7a6 (2x) + : a5 (2x)2

1.2

......

a1 (2x)3 + ...

7.6.5
1.2.3
=a7 - 7a6 (2x) +21a5 (2x)2 - 35a1 (2x)3 +35a3 (2x)

– 21a2 (2x)5+7a (2x)6 — (2x)7

=a7-14a6x+84a5x2-280a4x3+560a3x4

-672a2x5+448ax6 – 128x7.

384. In the expansion of (x+a)", the coefficient of the second term is "C1; of the third term is "C2; of the fourth term is "C'; and so on; the suffix in each term being one less than the number of the term to which it applies; hence "C, is the coefficient of the (r+1)th term. This is called the general term, because by giving to r different numerical values any of the coefficients may be found from "C,; and by giving to x and a their appropriate indices any assigned term may be obtained. Thus the (r+1)th term may be written

n (n-1) (n-2)...(n−r+1) "Can-ra", or

*

xn-ra".

In applying this formula to any particular case, it should be observed that the index of a is the same as the suffix of C, and that the sum of the indices of x and a is n.

Example 1. Find the fifth term of (a+2x3)17.
The required term

=17C1a13 (2x3)1

17.16.15.14

1.2.3.4
=38080a13x12.

Example 2. Find the fourteenth term of (3-a)15.
The required term =15C13(3)2(-a) 13
= 15 C2 × (−9a13)
=-945a13.

(1+x)"=1+"C1x+"C2x2+

=1+nx+

385. The simplest form of the binomial theorem is the expansion of (1+x)". This is obtained from the general formula of Art. 382, by writing 1 in the place of x, and x in the place of a. Thus

+"Crx2+

+nChưa

n(n-1)

1.2

x 16a13x12

...

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[Art. 370.]

...

+xn;

xr.

the general term being (n-1) (n-2)...... ( n − r+1) ̧.

r

386. The expansion of a binomial may always be made to depend upon the case in which the first term is unity; thus

y

(x + y)" = { x ( 1 + 2 }) } " =x" (1+2)", where z=

x

* Called "factorial " and meaning the continued product of all the integers from 1 to r. Thus 4 1.2.3.4 = 24.

Example. Find the coefficient of x16 in the expansion of (x2 – 2x)1o.

10

We have

( _ 2x)10= 0 '(1-2)

and, since x20 multiplies every term in the expansion of

(1-2)10,

we have in this expansion to seek the coefficient of the term which contains

1
4

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Assume that these laws being any positive integer.

PROOF BY MATHEMATICAL INDUCTION.

387. By actual multiplication we obtain the following identities:

(a+x)2=a2+2ax+x2,
(a+x)3=a3+3a2x+3ax2+x3,

(a+x)1=a1+4a3x+6a2x2+4ax3+xa.

Selecting any one of these and rewriting so as to exhibit the laws of formation of exponents and coefficients, we have

4

4.3

(a+x)=a+a11x+ a4-2x2+

4.3.2
1.2.3

a4-8x3

1.2

Then (a+x)"=a2+nan¬1x+

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4.3.2.1

+

aox4(a°=1 Art. 238).

1.2.3.4

of formation hold for (a+x)", n

n(n-1)

1.2

-an-3x3

n(n-1)(n-2) +

1.2.3

× 16

n(n+1)

1.2

n(n+1)(n−1)
1.2.3

an-2x2

.(1).

Multiplying each side of the identity by (a+x) and combining terms, we obtain,

(a+x)n+1=an+1+(n+1)a"x+

・an-1x2

+

an-2x3

.(2).

It will be seen that n in (1) is, in every instance, replaced by (n+1) in (2). Hence if the theorem be true for any value n, it will be true for the next higher value. We have shown

of

by multiplication that the theorem is true when n successively equals 2, 3, and 4; hence it is true when n=5, and so on indefinitely. The theorem is therefore true for all positive integral values of n.

388. In the expansion

(a+x)n=an+nan−1x+

n(n-1)
1.2

We observe that in any term

(1) The exponent of x, the second term of the binomial, is one less than the number of the term from the first.

(2) The sum of the exponents is n.

(3) The last factor of the denominator is the same as the exponent of the second term of the binomial.

(4) The numerator contains as many factors as the denominator, the first being n, and each succeeding one decreasing by 1.

Hence the (r+1)th or general term of (a+x)" is

n(n-1)... for r factors
1.2.3...r

6

7. (2-2).

Example. Find the 6th term in the expansion of (2a+b)1o.
Here n=10, and r+1=6.

We have

-an-2x2...

(2a)565

10.9.8.7.6
1.2.3.4.5
=252 (2)5α5b5 = 8064a5b5.

Expand the following binomials :

1. (x+2).

4. (a-x).

Write down and simplify:

10. The 4th term of (1+x)12. 12. The 5th term of (a—5b)7.

-an-rxr.

3.2.7.6

1

EXAMPLES XXXIX. a.

2. (x+3)5.

5. (1-2y)5.

8. (a–3)7.

(2a)5b5

(;

14. The 7th term of 1

16. Find the value of (x−√√3)1+(x+ √3)*.

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11.

The 6th term of (2-y).

13. The 15th term of (2x-1)17.

10

1) 15. The 6th term of (3x+

(3x+2)

9

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