Advanced book on Mathematics Olympiad

432 Algebra

···

an− 1 − 2 an+an+ 1 =bn

in the unknownsa 1 ,a 2 ,...,an. To determineakfor somek, we multiply the first equation
by 1, the second by 2, the third by 3, and so on up to the(k− 1 )st, which we multiply by
k−1, then add them up to obtain

−kak− 1 +(k− 1 )ak=

∑

j<k

jbj.

Working backward, we multiply the last equation by by 1, the next-to-last by 2, and so
on up to the(k+ 1 )st, which we multiply byn−k, then add these equations to obtain

−(n−k+ 1 )ak+ 1 +(n−k)ak=

∑

j>k

(n−j+ 1 )bj.

We now have a system of three equations,

−kak− 1 +(k− 1 )ak=

∑

j<k

jbj,

ak− 1 − 2 ak+ak+ 1 =bk,

−(n−k+ 1 )ak+ 1 +(n−k)ak=

∑

j>k

(n−j+ 1 )bj

in the unknownsak− 1 ,ak,ak+ 1. Eliminatingak− 1 andak+ 1 , we obtain
(
k− 1
k

− 2 +

n−k

n−k+ 1

)

ak=bk+

1

k

∑

j<k

jbj+

1

n−k+ 1

∑

j>k

(n−j+ 1 )bj.

Taking absolute values and using the triangle inequality and the fact that|bj|≤1, for all
j, we obtain
∣
∣∣
∣

−n− 1

k(n−k+ 1 )

∣

∣∣

∣|ak|≤^1 +

1

k

∑

j<k

j+

1

n−k+ 1

∑

j>k

(n−j+ 1 )

= 1 +

k− 1

2

+

n−k

2

=

n+ 1

2

.

Therefore,|ak|≤k(n−k+ 1 )/2, and the problem is solved.

240.The fact that the matrix is invertible is equivalent to the fact that the system of linear
equations

x 1

1

+

x 2

2

+···+

xn

n

= 0 ,