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17. Expand (V1 — 2+1)5–(V1–22-1)5. 18. Find the coefficient of 212 in (x2 + 2x) 10.

19. Find the coefficient of x in

20. Find the term independent of x in (2x2.

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21. Find the coefficient of x -20 in

In the expansion of (1+x)" the coefficients of terms equidistant from the beginning and end are equal.

The coefficient of the (r+1)th term from the beginning is ncc

The (r+1)th term from the end has n+1-(r+1), or n-. terms before it; therefore counting from the beginning it is the (n-r+1)th term, and its coefficient is nCn-m, which has been shown to be equal to Cp [Art. 370.] Hence the proposition follows.

390. Greatest coefficient. To find the greatest coefficient in the expansion of (1+x)".

The coefficient of the general term of (1+x)" is "Cr; and we have only to find for what value of r this is greatest.

By Art. 375, when n is even, the greatest coefficient is nCn; when n is odd, it is "Cn-1, or "Cn+1; these coefficients being equal.

391. Greatest term. To find the greatest term in the expansion of (x+a)". We have (x+a)"=2"

; therefore, since za multiplies every term in (1+ it will be sufficient to find the greatest term in this latter expansion.

Let the pth and (r+1)th be any two consecutive terms. The (r+1) th term is obtained by multiplying the pth term by n-r+1 a

2+1 that

[Art. 384.]

2

n+1 The factor -1 decreases as go increases; hence the

1 (3+1)th term is not always greater than the poth term, but only

(n+1 until

becomes equal to 1, or less than 1.

a

(142 - 1)

2

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(n+1) a If be an integer, denoto it by p; then if p=P

the xta multiplying factor becomes 1, and the (p+1)th term is equal to the pth; and these are greater than any other term. If (n+1) a be not an integer, denote its integral part by q;

xta then the greatest value of r consistent with (1) is q; hence the (2+1)th term is the greatest.

Since we are only concerned with the numerically greatest term, the investigation will be the same for (x a)"; therefore in any numerical example it is unnecessary to consider the sign of the second term of the binomial. Also it will be found best to work each example independently of the general formula.

Example. Find the greatest term in the expansion of (1+4x)”,

1 when x has the value

3.

Denote the pth and (r +1)\n terms by T, and Tr+1 respectively; then 8 - 7+1

9-1

4 Trt1

.4 x Tr

X

3

x Tri

r

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The greatest value of r consistent with this is 5; hence the greatest term is the sixth, and its value

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392. To find the sum of the coefficients in the expansion of (1+x)".

In the identity (1+x)=1+07.0+Cox2 + C3.23 + ... + CXT, put x=1; thus

2n=1+C1+C2+3+...+Cn

=sum of the coefficients. COR,

C1+C2+Cz+...+Cn=2n – 1; that is, the total number of combinations of n things taking some or all of them at a time is 2n – 1.

-1; thus

393. To prove that in the expansion of (1+x)", the sum of the coefficients of the odd terms is equal to the sum of the coefficients of the even terms.

In the identity (1+x)"=1+C7x+Cox? +C223+ ... + Chol", put x=

0=1-C1+C2-Cz+C4 - Cg + ......;

::1+C+Ce+ ......=C, +Cz+Co+ ....... 394. The Binomial Theorem may also be applied to expand expressions which contain more than two terms.

Example. Find the expansion of (x2 + 2x – 1)3.
Regarding 2x – 1 as a single term, the expansion

=(x2)3 + 3 (x?)? (2x – 1) + 3x2 (2x – 1)2 + (2x - 1)3
= X6 + 6x5 + 9x4 – 4x3 – 9x2 + 6x – 1, on reduction,

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x3 + ...

395. For a full discussion of the Binomial Theorem when the index is not restricted to positive integral values the student is referred to Chap. XLVII. It is there shown that when x is less than unity, the formula

n (12 – 1) n (n 1) (1–2) (1+x)"=1+na+

1.2

1.2.3 is true for any value of n.

When n is negative or fractional the number of terms in the expansion is unlimited, but in any particular case we may write down as many terms as we please, or we may find the coefficient of any assigned term.

+

Example 1. Expand (1+x) –3 to four terms.

(-3) (- 3 - 1)

(-3)(-3-1)(-3-2) +.... (1+x)-3=1+(-3) x+

1.2

1.2.3
3.4 3.4.5
=1-3. +

1.2 1.2.3
=1- 3.c +62 – 10x3 + ...

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203 + ...

3

3

Example 2. Expand (4+ 3x)2 to four terms.

3 3x\2

3x (4+ 3x)2=42 (1+

4

1+

4

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396. In finding the general term we must now use the formula

n (n − 1)(n-2)......(n-r+1) go

L2 written in full; for the symbol "C, cannot be employed when n is fractional or negative.

Example 1. Find the general term in the expansion of (1+x)]. 1/1

r+ 22 The (r+1)th term:

1 (1-1) (-2

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The number of factors in the numerator is r, and -1 of these are negative; therefore, by taking - 1 out of each of these negative factors, we may write the above expression (-1)=-11.3.5 ...... (27 – 3)

2cm.

Example 2. Find the general term in the expansion of (1 – 2) – 3.

(-3) (-4) (-5) ...... (-3 – +1) The (r + 1)th term=

( - x)"
Ľ
3.4.5 ...... (r+2)
=(-1)"

(-1)" x1

xr

....1

3.4.5 ...... (r+2) (2r

1.2.3 (r+1) (r+2)

1.2 by removing like factors from the numerator and denominator.

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397. The following example illustrates a useful application of the Binomial Theorem.

Example Find the cube root of 126 to 5 places of decimals.

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In the following expansions find which is the greatest term: 1. (x+y) 17 when X=4, y=3.

2. (28 - y)28 when .x= -9, y=4.

2 3. (1+2)4 when w=

3

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