Combinatorics and Probability 749Integrating by parts, we obtainIk=∫ π 20(2 sinθ)^2 k−^1 (2 sinθ)dθ=(2 sinθ)^2 k(−2 cosθ)|π 2
0 +∫ π 20( 2 k− 1 )(2 sinθ)^2 k−^2 4 cos^2 θdθ=( 2 k− 1 )∫ π 20(2 sinθ)^2 k−^2 ( 4 −4 sin^2 θ)dθ= 4 ( 2 k− 1 )Ik− 1 −( 2 k− 1 )Ik.HenceIk=^4 kk−^2 Ik− 1 ,k≥1. Comparing this with
(
2 k
k)
=
( 2 k)( 2 k− 1 )( 2 k− 2 )!
k^2 ((k− 1 )!)^2=
4 k− 2
k(
2 k− 2
k)
,
we see that it remains to check the equality^2 πI 0 =1, and that is obvious.
860.We computeA^2 =
⎛
⎜
⎜⎜
⎜⎜
⎝
123 ··· n
012 ···n− 1
001 ···n− 2
..
...
.
..
.
... ..
.
000 ··· 1
⎞
⎟
⎟⎟
⎟⎟
⎠
=
⎛
⎜
⎜⎜
⎜⎜
⎝
( 1
1)( 2
1)( 3
1)
···
(n
1)
0
( 1
1)( 2
1)
···
(n− 1
1)
00
( 1
1)
···
(n− 2
1)
..
.
..
.
..
.
... ..
.
000 ···
( 1
1)
⎞
⎟
⎟⎟
⎟⎟
⎠
.
Also,
A^3 =
⎛
⎜
⎜⎜
⎜⎜
⎝
( 2
2)( 3
2)( 4
2)
···
(n+ 1
2)
0
( 2
2)( 3
2)
···
(n
2)
00
( 2
2)
···
(n− 1
2)
..
.
..
.
..
.
... ..
.
000 ···
( 2
2)
⎞
⎟
⎟⎟
⎟⎟
⎠
.
In general,Ak=⎛
⎜
⎜⎜
⎜⎜
⎜
⎝
(k− 1
k− 1)(k
k− 1)(k+ 1
k− 1)
···
(k+n− 2
k− 1)
0
(k− 1
k− 1)(k
k− 1)
···
(k+n− 3
k− 1)
00
(k− 1
k− 1)
···
(k+n− 4
k− 1)
..
.
..
.
..
.
... ..
.
000 ···
(k− 1
k− 1