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To find the value of the series (2), we proceed thus:

€1x+2c2x2+3¤ ̧x3+ + ncn xn

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If we multiply together the two series on the left-hand sides of (3) and (4), we see that in the product the term independent of x is the series (2); hence

1\n-1

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In the following expansions find which is the greatest term:

1. (x-y)30 when x=11, y=4.

2. (2x-3y)28 when x=

=9, y=4.

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In the following expansions find the value of the greatest term:

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7. Shew that the coefficient of the middle term of (1+x)2n is equal to the sum of the coefficients of the two middle terms of (1+x)2n-1.

8. If A be the sum of the odd terms and B the sum of the even terms in the expansion of (x+a)”, prove that A2 – B2= (x2 — a2)n. 9. The 2nd, 3rd, 4th terms in the expansion of (x+y)" are 240, 720, 1080 respectively; find x, y, n.

10. Find the expansion of (1+2x − x2)1.

11. Find the expansion of (3x2 −2ax+3a2)3. 12. Find the 7th term from the end in (x+a)n.

13. Find the (p+2)th term from the end in

2n

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14. In the expansion of (1+x)43 the coefficients of the (2r+1)th and the (r+2)th terms are equal; find r.

15. Find the relation between r and n in order that the coefficients of the 37th and (r+2)th terms of (1+x)2n may be equal.

16. Shew that the middle term in the expansion of (1+x)2n is

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c, denote the coefficients in the expansion of (1+x)",

..+ncn=n. 2n-1

2n+1 1

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......

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=

Со

C1

C2

Cn-1

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n+1

n (n+1)

2

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CHAPTER XIV.

BINOMIAL THEOREM. ANY INDEX.

IN the last chapter we investigated the Binomial Theorem when the index was any positive integer; we shall now consider whether the formulæ there obtained hold in the case of negative and fractional values of the index.

Since, by Art. 167, every binomial may be reduced to one common type, it will be sufficient to confine our attention to binomials of the form (1 + x)”.

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and in each of these series the number of terms is unlimited.

In these cases we have by independent processes obtained an expansion for each of the expressions (1 + x) and (1 + x)-*. shall presently prove that they are only particular cases of the general formula for the expansion of (1+x)", where n is any rational quantity.

This formula was discovered by Newton.

178. Suppose we have two expressions arranged in ascending powers of x, such as

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The product of these two expressions will be a series in ascending powers of x; denote it by

1 + Ax+Bx2 + Сx2 + Dx1 +

......

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then it is clear that A, B, C, are functions of m and n, and therefore the actual values of A, B, C, ...... in any particular case will depend upon the values of m and n in that case.

But

the way in which the coefficients of the powers of x in (1) and (2) combine to give A, B, C, ...... is quite independent of m and n; in other words, whatever values m and n may have, A, B, C, ... preserve the same invariable form. the form of A, B, C, ...... for any that A, B, C, will have the and n.

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If therefore we can determine value of m and n, we conclude same form for all values of m

The principle here explained is often referred to as an example of "the permanence of equivalent forms;" in the present case we have only to recognise the fact that in any algebraical product the form of the result will be the same whether the quantities involved are whole numbers, or fractions; positive, or negative.

We shall make use of this principle in the general proof of the Binomial Theorem for any index. The proof which we give is due to Euler.

179. To prove the Binomial Theorem when the index is a positive fraction.

Whatever be the value of m, positive or negative, integral or fractional, let the symbol f(m) stand for the series

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If we multiply these two series together the product will be another series in ascending powers of x, whose coefficients will be unaltered in form whatever m and n may be.

To determine this invariable form of the product we may give to m and n any values that are most convenient; for this purpose suppose that m and n are positive integers. In this case f(m) is the expanded form of (1 + x)", and f(n) is the expanded form of (1+x)"; and therefore

ƒ (m) × ƒ (n) = (1 + x)TM × (1 + x)” = (1+x)m+n,

but when m and n are positive integers the expansion of (1 + x)TM

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m+n

This then is the form of the product of f(m) ×f(n) in all cases, whatever the values of m and n may be; and in agreement with our previous notation it may be denoted by ƒ(m +n); therefore for all values of m and n

Also

f(m) ׃(n) =ƒ (m+n).

f(m) ׃(n) ׃ (p) =ƒ (m + n) × ƒ (p)

=f(m+n+p), similarly.

Proceeding in this way we may shew that

ƒ(m)׃(n) ׃(p)...to k factors = f (m +n+p+...to k terms).

Let each of these quantities m, n, p,

where h and k are positive integers;

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but since h is a positive integer, ƒ (h) = (1 + x)" ;

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