Advanced book on Mathematics Olympiad

424 Algebra

Let

α=max

i

⎛

⎝

∑n

j= 1

|aij|

⎞

⎠< 1.

Then

∑

k

∣

∣∣

∣∣

∣

∑

j

aijajk

∣

∣∣

∣∣

∣

≤

∑

j,k

|aijajk|=

∑

j

(

|aij|

∑

k

|ajk|

)

≤α

∑

j

|aij|≤α^2.

Inductively we obtain that the entriesaij(n)ofAnsatisfy

∑

j|aij(n)|<α

nfor alli.

Because the geometric series 1+α+α^2 +α^3 +···converges, so doesIn+A+A^2 +
A^3 +···.And the sum of this series is the inverse ofIn−A.
(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)

225.The trick is to computeA^2. The elements on the diagonal are

∑n

k= 1

sin^2 kmα, m= 1 , 2 ,...,n,

which are all nonzero. Off the diagonal, the(m, j )th entry is equal to

∑n

k= 1

sinkmαsinkj α=

1

2

[n

∑

k= 1

cosk(m−j)α−

∑n

k= 1

k(m+j)α

]

.

We are led to the computation of two sums of the form

∑n
k= 1 coskx. This is done as
follows:

∑n

k= 1

coskx=

1

2 sinx 2

∑n

k= 1

sin

x

2

coskx=

1

2 sinx 2

∑n

k= 1

[

sin

(

k+

1

2

)

x−sin

(

k−

1

2

x

)]

.

The sum telescopes, and we obtain

∑n

k= 1

coskx=

sin

(

n+^12

)

x

2 sinx 2

−

1

2

.

Note that forx=(m±j)α=(mn±+j)π 1 ,

sin

(

n+

1

2

)

x=sin

(

(m±j)π−

x

2

)

=(− 1 )m+j+^1 sin

x

2

.

Hence