2 + 2 (a − 1) + 2 + 2 (a − 1) +

α 1 2a - 1 4a2 - 5a + 1

8a2 - 8a+ 1

8a-4

[13 and 14 are connected, because a2- a = (a1)2+ a − 1.]

20.

and

(240)2

2(2111)

21.

and

(273)

2 (2885)

19. 185, 15; 119, 81; 53, 147. 20. 28 crowns, 20 half-crowns. 21. When n is even, the common difference is 2; when n is odd, the common difference may be 1 or 2.

22. 245.

25. Ascribe to y succes

sively the values 1, 2,... 8; and in each case find the corresponding values of x and z. 26. x=1+3t, y=51-7t, z= = 63+13t. 27. Allowing a zero, there are 15 solutions; excluding it, there are 14. The solutions are found from 100-t half-crowns, 6t 28. Allowing zeros,

shillings, and 100 - 7t sixpences.

4 solutions; excluding them, 2. The solutions are found from

4-t guineas, 5t crowns, and 12– 4t shillings.

29. 6 crowns,

reckoned from either of the common ends.

39. We must solve 5x+4y+32=20: the accompanying table

12. 1+px+p(p-1)x+(p3- 2p3+1)x+p(p3- 3p2+p+2)x*......

11. /1/

a3

1 + x

2.

3.

1 – 10x + 211⁄22; general term 2(7x)” — (3x)”.

4.

96

1

; 1.

1 + n

1

3

11

2 (n + 2) (n + 3) 4 (n + 1) (n + 2) (n + 3) (n + 4)
+4) 96'

where A1, A,, A do not contain v.

......

Now change v into xv; thus we can infer that

19. 460.

m

Now equate the coefficients of the same powers of v on the two sides.

Expand each term of the last line by the Binomial Theorem and then equate the coefficients of x" on the two sides.

LI. 8. 2x is> or <x+1 according as x is> or < 1.

16. This depends on the sign of (a - b) (b − c) (c − a). 22 and 24 depend on Art. 676.

23. As many of the fol

lowing inequalities as may be required will be found to hold;

2(n-1)>n, 3 (n-2) >n, ......; then by multiplication the

result is obtained.

25. may be deduced from Ex. 23.

29. See Ex. 3 of Chap. xxv.

31. Multiply up; then use

Art. 676.
Binomial Theorem; the series will be convergent.
then have to shew that

32. Put 1-ab, and expand (1-6) by the

LII. 2. 66.

5. 2°. (823).

13

3. 3.5o. 41°.

1;

We shall

4. 2o. 3o, 5o.

12. suppose n to lie between m3 and (m + 1)2; 19. n-n+1, is greater than

then n― ab = (m2 + m − n)3. (n-1) and less than n3.

then n3 = (m-1) (m + 1).

3

20. Suppose, if possible, n3 + 1 = m3 ; Now no factor, except 2, can divide

both m 1 and m+1, and 2 cannot here divide them, for n is odd. Hence m −1 and m+1 must both be perfect cubes; but this is impossible; for the difference of two cubes cannot be so small as 2. 35, 36, 37, 38. These all depend on Fermat's

53 and 54 must be solved by trial; the answer to 53 is 2. 3o. 5,

and the answer to 54 is 23. 33. 5.7.

y=2.5.7.t.

57. x=2.5.7.0,

is, and of neither winning nor losing is ; D's chance of winning is, and of neither winning nor losing is ; B and C have each the chance of winning, of losing, of neither. simply, A's chance of winning is, B's and C's, and D's &, if

we suppose that one of the boats must win.

Or more