m-2 m-2 3 m 3 m m m 1 5 5 m m 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 +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 1-27 6. Solve the following equations : - 69 Result =84 :87. (3) 2cx' – abx + 2abd=4cd.c. Result <= or 2d. (4) (x + y) – 2* = 65) (i). 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; ข therefore y=2 or .18, whence easily x=7, -29; 2=4, 49 13 2 202 + 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.). 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 + gallons 80 100 ca of brandy, that is, C- gallons; and therefore after 400 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 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 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. n 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 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 n n = 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. |