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m-2

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3 m

3 m

m

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if only a " + y is so. Now, a" + you is known to be

+ divisible by amryw without remainder; therefore so is

+" x"+y", &c.; and therefore so is cm +y”, that is, x + y,

2, , if only m is an odd number. The three last terms of the quotient required are

+

Ry and the quotient will contain 2n + 1 terms.

2 2n-2 2n+1 2n+1

27-1 2n+1 2n+1 C

2n 2n+1

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+y

5. Find the fraction in its lowest terms which is the square root of

1+ 4x+ – 2x – 12x + 9x2

1–4x3 + 6x - 4x + acas
1+ 3.2
Result

1-27

6. Solve the following equations :
(1) (2 - 4) (2 – 8) (2 – 10) = (x – 6) (– 7) (x – 9).

- 69 Result =84

:87.
(2) 53* – 14x = 3. Result x=3 or – š.

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(3) 2cx' abx + 2abd=4cd.c. Result <=

or 2d.

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(4) (x + y) – 2* = 65) (i).
2– (y +z)' = 13} (ii).

2+2- y= 9) (iii). To solve this last equation, subtract (ii) from (i) and

26 we get 2y (2 + y + 2) = 52; that is, x+y+% ; and

y

therefore from this equation and (iii) we obtain

26 2y =

-9;

therefore

y=2 or .18, whence easily x=7, -29; 2=4, 49

13

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202 +

7. Form the equation whose roots are – 5+6 V(-1).

If x2 - px +q=0 and 3* 2,2+2, = 0 have a common root, determine it; and in that case shew that

(9-9.)* = (p-p) (2,2-29.).
The equation whose roots are – 5+6 V(-1) is

Da + 61=0. If the equations wc-px+q=0 and a* - 2,2 +9,= 0 have a common root, this root satisfying each eqnation will satisfy their difference, that is, the equation

-px +9+PX-9.=0; therefore the common root = x=

9.-9

P.-P Substituting this value of x in either of the given equations, we obtain the identity required.

8. A bill of £63. 58. was paid in sovereigns and half-crowns, and the number of coins used in the payment was 100. How many sovereigns were used, and how many half-crowns ?

Let x=the number of sovereigns paid, y=the number of half-crowns; then we have the two equations

20x + 2 y = 1265,

x+y=100, from which we obtain x = 58, y = 42, and therefore the

=

number of sovereigns paid is 58 and the number of halfcrowns 12.

9. A cask (A) is filled with 50 gallons of water, and a cask (B) with 40 gallons of brandy; (C) gallons are drawn from each cask, mixed, and replaced. The same operation is repeated. Find C when there are 87 gallons of brandy in (A) after the second displacement.

After first displacement A contains 50 – 10 gallons of water and įC gallons of brandy, B similarly contains 40-1C gallons of brandy and IC of water.

с Now on second drawing we take (50—1C) gallons

50

02 of brandy from A and of brandy, and from B we

100 c

C2 (–

of water. 40

80

C2 02
Therefore the mixture contains -

+ gallons 80

100

ca of brandy, that is, C- gallons; and therefore after

400
C2

C2 replacement A contains 10.

+ 10

gallons 100

800

9 C2 of brandy; therefore we have C- 7, from which

800 we find C=° or 10, the latter of which is the only admissible value.

take (40 – 4C) of brandy and

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71

10. Find the sum of (n) terms of the series (a - b) + (a – 36) + &c. ... without assuming a formula for the sum of an arithmetic series, and apply the result to find the sum of 10 terms of 76 + 70 +64 + &c.

(a - b)+(a – 36) + (a – 56) + &c. to n terms =na nob.

The series 76 + 70 + 64 + &c. to 10 terms

= (79 – 3) +(79 – 9) + (79 – 15) + &c. to 10 terms = 790 – 300 = 490.

=

11. The sum of the first two terms of a geometric series is 12, and of the first three terms is 39; find the series, and determine if the condition is satisfied by more than one series. Find the sum of the infinite series

1+r+(1+6) p2 + (1 + b + 8%) 18 + &c., r and b being proper fractions. .

Let a, ar, ar?, &c. be the first three terms, &c. of the geometric series mentioned above, then we have

a+ar=12, and

a +ar+ar = 39, from which we find r=

3 or , and a=3 or 48; therefore the two series required are

3,9, 27, &c., 48, - 36, 27, &c. To sum the infinite series

1+r+ (1+6) 7* +(1+b+%) 18 + &c., this = (1+r+gol + &c.) + brl (1 +r+q* + &c.)

+b%(1+r+p+ &c.) +&c. 1+ br* + % 8 + &c. 1

br2 l 1-r

1-br 1-br + bra 1

1-br -7

49

1+

1

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12. Shew how to transfer a whole number from one scale of notation to another, and prove that if r be the radix of the scale, the number itself will be divisible by p+ 1 if the difference between the sum of the digits in

the odd places and the sum of the digits in the even places be divisible by r + 1.

Tod. Alg., Arts. 430 and 444.

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13. Find the number of different permutations of n letters taken altogether, when one letter occurs p times, and another q times in each permutation. .

Tod. Alg., Art. 498.

If the n letters contain only a and b and n be even, shew that the number of permutations will be greatest when p=q=in.

р
We have (2[n-p=n" (p-1 [n-(p-1);
P (

р

-; P

P therefore LP (n–p will decrease so long as <1,

-P and therefore to find when the number ( 2 (n – -p is

р least, that is, when ceases to decrease, we must put

р p =1, from which we obtain prin; therefore

; р 2 (n-p has its least value when p= in, and therefore

p (where q=n-p) will have its greatest value [2 ( when p=9= {n.

п

n

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n

14. Assuming the form of the binomial theorem, find the greatest term of (1 + x)" when n is a positive integer, and x is also positive.

Tod. Alg., Art. 520.

Apply the result to the determination of the greatest term of (1 + 5), and find the value of the greatest term.

The fifth and sixth terms are equal to 2375, and these are each greater than any other term.

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