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and, integrating with respect to x, we have
n-i+1
(n-i + 1)(n-i + 2) ... (n-1)-

(n-1) (n-i+1) ... (n-1) (i-1)!

i(i-2)!

n!

Σ

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i! (n-1)!' which is the number of points scored by the player in i periods. Putting i = 1, 2, in, the sum of the results is 2" – , which is the total number of points that can be scored by the player. Now the whole number of ways in which the player can score is (n-1-ai)!

from di=1 to ai = n-i +1,
( -4 +1=a;)!
1

(n-1)!
3 x ir )
(i-2)!

(i-1)! (n-1)!' and in each of these there are i-1 failures, or (-1)

(n-1)!

in

(i-1)! (n-2)! all. Let i = 1, 2, n; then the number of failures

+ 3 (n— 1) (n-2) (n-3) + ... + (n-1). + 2 1.2

1.2.3 Now (1+x)n-1 1+ (n − 1) x+

(n-1) (n − 2) x2 + ... +in-1

1.2 Differentiating with respect to x, we have (n-1) (1+x)n-2 = 1.(n-1) +2

(n-1) (n − 2)

x+ ... + (n-1) xn-2.

1.2 Let x=1; therefore (n-1) 2n–2 1.(n-1) + 2

(n − 1) (» — 2). +(n-1).

1.2 This result is obtained on the supposition that the player finally scores. But he can finally fail as often as he can finally succeed, namely, 21-1 times. Therefore the whole number of points that can be scored by the non-player is (n-1) 21–2 + 2" – 1 = (n + 3) 2"-2 – 1. Hence their respective probabilities are

2n-1 (n+3) 21-2–1 (n+7) 21-2-2' (n+7) 2n-2-2

1

+

5621. (By D. EDWARDES.)—If a circle be drawn through the centre of the inscribed circle and the centres of any two escribed circles of a triangle, prove that its radius is double that of the circumscribed circle of the triangle.

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Solution by VINCENZO JACOBINI ; W. S. F. LONG, B.A.; and others. Sia ABC il triangolo, siano 0, 01, 02, 03 i centri del circolo inscritto e dei circoli ex-inscritti al me desimo; allora il circolo ABC sarà rispetto al triangolo 0,0,0, il circolo dei nove punti, come il punto O sarà rispetto allo stesso il punto di concorso delle altezze, quindi il raggio del circolo ABC sarà metà del raggio del circolo 0,0,03, ma quest' ultimo è eguale a quello dei circoli 00,02, 00,03, 00,03; come si vede facilmente osservando che: triangoli 0,0,03, 00,0., 00,03, 00,0g hanno lo stesso circolo dei nove punti, dunque i circoli 00 02, 00, 03, 00.03 hanno i raggio doppio di quello del circolo ABC.

136. CONIC CONSTRUCTIONS.

By E. J. LAWRENCE, M.A.

Take a frame ABCD, consisting of four links crossed as in the figure, in which AB = opposite link CD, and BC = opposite link AD. Suppose the latter pair of links the longer of the two. ' Then

(1) If a short link CD be absolutely fixed, and the rest of the frame moved about it in the plane of the paper, the intersection P will describe an ellipse having C, D for foci, axis major AD, and the tan. gent to the curve at P passes through Q the intersection of AB and CD produced

(2) If a long link BC be absolutely fixed and the frame moved as before, Q, the intersection of the short links produced, will describe a hyperbola whose foci are B, C, its axis major AB, and tangent to the curve at Q passes through P. The asymptotes will be parallel to the two directions in which AB and CD lie parallel to one another.

5573. (By Professor Monck, M.A.)—The three edges and the diagonal of a rectangular parallelepiped are integer numbers ; show how to obtain a series of parallelepipeds possessing the same quality.

Solution by the PROPOSER. Let a, b, c be the edges of the parallelepiped, and d the diagonal. Another can be formed by taking for the new edges a+b+d, a +c+d, and b+c+d, and for the new diagonal a + b +c+ 2d; and the process can

+

be repeated ad infinitum. (Any of the edges can be taken as negative with a similar result.) For (a+b+d)2 + (a +c+d)2 + (b+c+d)2

2a2 +262 +2c2 + 2ab + 2bc + 2ac +4 (a + b + c) d + 3d, or (since a2 + b2 + c = d2)

- a? +62 +62 + 2ab + 2bc + 2ac +2 (a + b + c)(2d) + 4 d2

(a + b + c)2 +2 (a + b + c) (20)+(22)2 = (a + b +c+ 2d)2 The simplest parallelepiped of the kind is that whose edges are 1, 2, and 2, and diagonal 3; next comes that whose edges are 2, 3, and 6, and diagonal 7, which is derived from the former by the above rule, taking one of the 2's as negative.

5436. (By Dr. Booth, F.R.S.)-Express 3 (sec A) and 3 (cosec A) in terms of the radii of circles connected with the given triangle ABC.

Solution by the EDITOR. From Booth's Geometrical Methods (Vol. II., pp. 294 (e), (f), 295 (6)], sin A sin B + sin B sin C + sin C sin A

92 +82 + 4Rr

.(1),

4R2 cos A cos B+cos B cos C +cos C cos A =

gol + 92 -4R2

(2);

4R2
18
sin A sin B sin C

cos A cos B cos C

р

..(3, 4). 2R2'

2R Dividing (1) by (3), and (2) by (4), and bearing in mind that 7&s2 = rr17273, 52 = r2r3 + 73?i tror= 4R2 + 4 Rr + g? + 2Rp,

9.2 + 32 + 4Rr we have

2R2 + 4Rr + po? + Rp 2rs

( (rrirara) and

72 +82-4R 2Rr + po2 +

- Rp. Σ

2Rp

Rp

(sin A) (COSA)

5549. (By A. W. Panton, M.A.)- A plane cuts a spheroid of revolution, and makes an angle a with the axis ; prove that the eccentricity of the section is equal to e cos a, where e is the eccentricity of a section through the axis.

Solution by R. E. RILEY, B.A.; R. F. Davis, B.A. ; and others. It will easily be seen that if a, b be the axes of the principal section, and d', 6 those of an oblique section through the centre, then b'=b, and a' = that radius vector of the principal section making an angle a with the axis = 6(1 - p2 cos? a)*; therefore &c.

23

+

+

+ ...,

5492. (By M. HERMITE.)-Soit

x? F(x) = 1+

n+1 ' (n + 1)(n + 2) (n + 1) (n + 2)(n + 3) on demande de démontrer qu'on a F)F(-) 1

x2

x4 na

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

+

+

I. Solution by J. J. WALKER, M.A.; R. E. RILEY, B.A.; and others.

+

+

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) =

a'

du Assume F (2) F(-2) = U, then = F'(x) F(-X)-F(x) F'(-x),

da.
or, by (1), a - * {F(x) + F(-x)} – 2ņu, whence we have
du

x2
U = 2nx21 - 1
(1+

...). dx

(n + 1) (n + 2)(n+1) ... (n + 4) Integrating, we obtain 22n u = c + x2n +

(n + 1) (n + 2) (n + 1) (n + 2)2 (n + 3) (n + 4) By putting x=0 the value of the arbitrary constant is found to be zero ; and dividing the last equation by nox?n, the result is as in the Question.

x2n

+ 2nx2n-1

+

+

+

nx2n + 2

nx2n + 4

+

+..

xn+1

x + 2

+

+

+...

II. Solution by J. HAMMOND, M.A.; Professor BATTAGLINI; and others. x" F (r)

o, suppose;
r (n+1) r(n+1)' r (n + 2)'r(n+3)
"F(-x)

t, suppose ;
r(n+1) r(n+1) r(n + 2) 'T(n + 3)
do

which, multiplied by e-x and integrated, gives dac r(n)'

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In this result put ux for y, and vx for z, and divide by 22n, then
F(x) F(-X)

fit (- 0) (uv)a-1 du dv.

Differentiate this 2m times with respect to x, and put x = 0 to get the 2mth coefficient, then F(a) F(-X)

:S

(1—02m (10)n-1 du do

x2m

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The general value of this definite integral for any value of m I have found to be

B (2m +1, n), but as the working is not elegant I have not

mtn
given it.
Assuming this, then the general coefficient is
B (2m + 1, n)

r(n)
(m+n) 2m ! (m + n) r(2m +n +1)

1
n (n + 1) ... (n + m)* (n + i + 1) ... (n + 2m)

III. Solution by ROBERT RAWSON ; Professor MANNHEIM ; and others.
Let
F(x) = 40 + aj x + 2g 22 + zx3 +

.(1), therefore F(-X) = 2o— Q7x + 29x2 – Az x3 +

(2); therefore F(x) F(-x) = A0 + A22+ A42* + ...

.(3), where An = a*, A, 2a,do-az, A4 = 2a4do – 2aza, + az*,

Ав 228%0 – 205a, + 204Qz_az,
Ag = 2agao - 2272+ 2agaz – 2agaz + aš?,

>(4), :

A2m 2a2m. 20—2a2m-1. 2, + 2a2m-2. ag-202m-3. Ag + tam the plus sign being used when (m) is even, and the minus sign when (m) is odd. By the conditions of the question, we have

1 do = 1,

an

,
an
Ag
Az =

...... a2m =
n+2'

n + 2m Substituting these values in (4), we have

n + 2

па. A. 1,

= 12 ( 2 02

n +1,

02m - 1

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n +3'

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