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There are two cases to consider:

1. When n is even, the number of terms in the expansion is odd; and the middle term is greater than any other term. For example, when ʼn is even, the coefficients of the development of (1 + x) are 1,6 15, 20, 15, 6, 1;

hence, the coefficient 20 is the largest.

2. When n is odd, the number of the terms is even, and the two coefficients equally distant from the extremities of the development are the greatest. For example, the development of (1+x)' has the coefficients

1, 9, 36, 84, 126, 126, 84, 36, 9, 1,

of which the two coefficients 126, 126, are the largest.

627. The preceding discussion gives a property of combinations worthy of notice. Suppose that one desires to know, for example, in what way six objects must be combined in order to obtain the greatest number of combinations. It is evident that the six objects should be taken three at a time, because the coefficients of the development of (1+r)", beginning with the second, are the number of combinations of six objects taken respectively one, two, three, etc., at a time; the largest coefficient being the fourth. Whence it follows that the number of combinations of six objects taken three at a time is the greatest of all possible combinations of six objects. Similarly, in case of nine objects, the greatest number of combinations is obtained on taking four or five at a time.

Suppose that n is odd, and equal to 2p+ 1; then

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Now for all values of r up to p inclusive, "+1

is greater than r,

1626, (8), but if r = p + 1, the multiplying factor

n-r+1 2p+1-p−1+1

=

[2626, (7)]

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Therefore, the number of combinations is the greatest in case

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628. Suppose x = 1 in equation (6), 624, for the expansion of

(1 + x)"; then,

(9) (1+1)" =2" = 1 + „C1 + n C2 + n C3 + ...

n 1

nСz+nC3+

2

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Therefore, it follows from (9) that the sum of the coefficients of the development of (x+a)" is

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It follows also that the total number of combinations which can be made with n objects by taking them in all possible ways, one at a time, two at a time, etc., is 2" — 1 (622).

629. If one puts a=1, z=1 in the development for (x-a)", (2) 623, he obtains

(1 − 1)” = 0 = 1 — nĘ1 + n C2 − n Cs+... ±nCni
— nC1+nC2−

whence it follows that

(11)

3

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Hence, when all possible combinations of n objects are formed, the number of combinations which can be made in case odd numbers of them are taken is greater by unity than the number of combinations which can be made when even numbers of them are taken. Let these two numbers of combinations be v and w; then by (9), 1628, (12) formula (11)

v + w = 2n

v

- w = 1;

ཇ་

whence, it follows, from (12) and (11), that

(13)

v = 2"-1,

w2-11.

For example, with 11 objects one can form in all 2"-1, that is, 2047, combinations. Of these combinations, there are 1024 which are composed of an even number and 1023 of an odd number of objects.

630. Summation of the Same Powers of Numbers which Form an Arithmetical Progression.

The solution of this problem is attained by another application of the binomial theorem.

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be n terms of an arithmetical progression; call d the common difference of the successive terms and represent the sum of the th powers of all the terms by S.

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Let the equations be added member to member, and the equal terms (a,d) and ag1, (a,+d)+1 and ag+1, . . . (an_1+d)r+1 and art, which occur respectively in the two members, be suppressed; on replacing ad by a,+nd, it follows that

(2) (a ̧+nd)r+'=a'+'+'+as++] "des,++"+1 d'$',+ndr+1.

2!

...

On putting in (2) r = 1, one obtains the value of the known sum S of the n terms of progression (1), since in this case series (1) consists of three terms; then

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If one puts r = 2 in (2), the value of S is found by means of S1, thus

2

(a ̧ + nd)3 = a3 + 3 d S2 + 3 d2 S1 + nd3. .

Hence, replacing S, by its value,

a{+3a}nd+3a ̧n2ð2+d3n3=a}+3dS,+3d3["{2a,+(n−1)d}]+nd3.

3n

3

(4) 3$=3a}n+3n2a ̧d+ d2n3- મા 2ad- n(n-1) d2-nd2

2

=3a}n+(3n3a,—3na,)d+ {n3 — 3 n(n − 1)—n }d2

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2

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S, are found by means of the

3

631. In particular, if one desires the sum of the first, second, etc., powers of the first n integral numbers, he has only to put a1 =1, and d=1 in equation (2) and then proceed as has just been described.

I. If

a=1, d=1,

=1;

(1 + n)2 = 12 + 1 + 1 §1 + n · 12,
1S,+n 12, whence S n(n+1).

II. When a = 1, d= 1, r = 2; then,

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=

2

3 n (n+1)

+n,

2

n(n+1)(2n+1).
3!

[2605, (1)]

1 and d=1

The results of I and II can be deduced by putting a, in equations (3) and (4).

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632. The sum of the squares and the sum of the cubes of the first n integral numbers may be obtained by direct methods, without using the binomial theorem.

1. In order to find the sum of the squares, consider the following table, which contains n - 1 columns and n - 1 rows:

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If the sum of the numbers of this table be taken by horizontal rows, it follows that their sum is

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1)2 n(n
+

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Reckoning the sum of the numbers of this table by vertical columns, one obtains the sum

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..

i. e., n[1+2+3+ . . . + (n − 1)] — [1+2+32+ ... (n−1)o]

..

+ 3 (n − 3) + . . . + (n − 1) (n − (n − 1));

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On equating expressions (1) and (2), one finds

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2. The sum of the cubes of the first n numbers may be found by a method due to M. Barbier. Construct a table by the multiplication of the first n numbers:

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The sum of the numbers of this table found by adding the columns by horizontal rows is the square of the sum of the first n numbers, for the sum of the first column is the sum of the first n numbers, which we call N, multiplied by 1; the sum of the numbers of the second column is Nx 2, and of the third column is N× 3, and so on. The entire sum is therefore

N(1+2+3+...+n)=N.

This sum may be formed in another way. Group the products in the following manner:

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